\displaystyle \textbf{Exercise 10(A)}

\displaystyle \textbf{Question 1: }\text{Which of the following pairs of triangles are congruent?}
\displaystyle \text{(a) }\triangle ABC\text{ and }\triangle DEF\text{ in which }BC=EF,\ AC=DF
\displaystyle \text{and }\angle C=\angle F.

\displaystyle \text{(b) }\triangle ABC\text{ and }\triangle PQR\text{ in which }AB=PQ,\ BC=QR
\displaystyle \text{and }\angle C=\angle R.

\displaystyle \text{(c) }\triangle ABC\text{ and }\triangle LMN\text{ in which }\angle A=\angle L=90^\circ,
\displaystyle AB=LM,\ \angle C=40^\circ\text{ and }\angle M=50^\circ.

\displaystyle \text{(d) }\triangle ABC\text{ and }\triangle DEF\text{ in which }\angle B=\angle E=90^\circ
\displaystyle \text{and }AC=DF.
\displaystyle \text{Answer:}
\displaystyle \text{(a) In }\triangle ABC\text{ and }\triangle DEF,\displaystyle BC=EF
\displaystyle AC=DF
\displaystyle \angle C=\angle F
\displaystyle \therefore \triangle ABC\cong\triangle DEF\qquad\text{(SAS congruency criterion).}

\displaystyle \text{(b) In }\triangle ABC\text{ and }\triangle PQR,\displaystyle AB=PQ
\displaystyle BC=QR
\displaystyle \angle C=\angle R
\displaystyle \text{Here, the equal angles are not included between the two corresponding sides.}
\displaystyle \text{Thus, the given information represents the SSA case, which is not a valid}
\displaystyle \text{criterion for congruency.}
\displaystyle \therefore \triangle ABC\text{ and }\triangle PQR\text{ are not necessarily congruent.}

\displaystyle \text{(c) In }\triangle ABC,\displaystyle \angle A=90^\circ\text{ and }\angle C=40^\circ
\displaystyle \therefore \angle B=180^\circ-(\angle A+\angle C)
\displaystyle =180^\circ-(90^\circ+40^\circ)
\displaystyle =50^\circ
\displaystyle \therefore \angle B=\angle M\qquad\text{(each }50^\circ\text{).}
\displaystyle \text{Also, }\angle A=\angle L=90^\circ\text{ and }AB=LM.
\displaystyle \therefore \triangle ABC\cong\triangle LMN\qquad\text{(AAS congruency criterion).}

\displaystyle \text{(d) In }\triangle ABC\text{ and }\triangle DEF,\displaystyle \angle B=\angle E=90^\circ
\displaystyle AC=DF
\displaystyle \text{Only one pair of corresponding angles and the hypotenuses are equal.}
\displaystyle \text{This information is insufficient to prove the congruency of the triangles.}
\displaystyle \therefore \triangle ABC\text{ and }\triangle DEF\text{ are not necessarily congruent.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In the given figure, }P\text{ is a point in the interior of }\angle ABC.
\displaystyle PL\perp BA\text{ and }PM\perp BC\text{ such that }PL=PM.
\displaystyle \text{Prove that }BP\text{ is the bisector of }\angle ABC. \displaystyle \text{Answer:}
\displaystyle \text{Given: }P\text{ is a point in the interior of }\angle ABC,\ PL\perp BA,
\displaystyle PM\perp BC\text{ and }PL=PM.
\displaystyle \text{To prove: }BP\text{ bisects }\angle ABC,\text{ i.e., }\angle ABP=\angle PBC.
\displaystyle \text{Proof: In right-angled }\triangle BLP\text{ and }\triangle BMP,
\displaystyle \angle BLP=\angle BMP=90^\circ
\displaystyle BP=BP\qquad\text{(common hypotenuse)}
\displaystyle PL=PM\qquad\text{(given)}
\displaystyle \therefore \triangle BLP\cong\triangle BMP\qquad\text{(RHS congruency criterion).}
\displaystyle \therefore \angle LBP=\angle MBP\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Since }L\text{ lies on }BA\text{ and }M\text{ lies on }BC,
\displaystyle \angle LBP=\angle ABP\text{ and }\angle MBP=\angle PBC.
\displaystyle \therefore \angle ABP=\angle PBC.
\displaystyle \therefore BP\text{ bisects }\angle ABC.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the given figure, the equal sides }BA\text{ and }CA\text{ of }\triangle ABC
\displaystyle \text{are produced to }Q\text{ and }P\text{ respectively, such that }AP=AQ. \displaystyle \text{Prove that }PB=QC.
\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ BA=CA,\text{ and }BA\text{ and }CA\text{ are produced to }Q
\displaystyle \text{and }P\text{ respectively, such that }AP=AQ.
\displaystyle \text{To prove: }PB=QC.
\displaystyle \text{Proof: In }\triangle APB\text{ and }\triangle AQC,
\displaystyle AP=AQ\qquad\text{(given)}
\displaystyle AB=AC\qquad\text{(given)}
\displaystyle \angle PAB=\angle QAC\qquad\text{(vertically opposite angles)}
\displaystyle \therefore \triangle APB\cong\triangle AQC\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore PB=QC\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In the given figure, median }AD\text{ of }\triangle ABC\text{ is produced. If }BL
\displaystyle \text{and }CM\text{ are perpendiculars drawn on }AD\text{ and }AD\text{ produced, prove that} \displaystyle BL=CM.
\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AD\text{ is the median of }BC\text{ and is produced to }M.
\displaystyle BL\perp AD\text{ and }CM\perp AD\text{ produced.}
\displaystyle \text{To prove: }BL=CM.
\displaystyle \text{Proof: In }\triangle BLD\text{ and }\triangle CMD,
\displaystyle BD=CD\qquad\text{(since }D\text{ is the midpoint of }BC\text{)}
\displaystyle \angle BLD=\angle CMD=90^\circ
\displaystyle \angle BDL=\angle CDM\qquad\text{(vertically opposite angles)}
\displaystyle \therefore \triangle BLD\cong\triangle CMD\qquad\text{(AAS congruency criterion).}
\displaystyle \therefore BL=CM\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the given figure, }M\text{ is the midpoint of }AB\text{ and }CD.
\displaystyle \text{Prove that }CA=BD\text{ and }CA\parallel BD. \displaystyle \text{Answer:}
\displaystyle \text{Given: }M\text{ is the midpoint of }AB\text{ and }CD,\text{ and }CA\text{ and }BD\text{ are joined.}
\displaystyle \text{To prove: }CA=BD\text{ and }CA\parallel BD.
\displaystyle \text{Proof: In }\triangle ACM\text{ and }\triangle BDM,
\displaystyle CM=DM\qquad\text{(since }M\text{ is the midpoint of }CD\text{)}
\displaystyle AM=BM\qquad\text{(since }M\text{ is the midpoint of }AB\text{)}
\displaystyle \angle AMC=\angle BMD\qquad\text{(vertically opposite angles)}
\displaystyle \therefore \triangle ACM\cong\triangle BDM\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore CA=BD\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Also, }\angle ACM=\angle BDM\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{These are alternate interior angles formed by the transversal }CD.
\displaystyle \therefore CA\parallel BD.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the given figure, }PA\perp AB,\ QB\perp AB\text{ and }PA=QB.
\displaystyle \text{If }PQ\text{ intersects }AB\text{ at }M,\text{ show that }M\text{ is the midpoint of both }AB
\displaystyle \text{and }PQ. \displaystyle \text{Answer:}
\displaystyle \text{Given: }PA\perp AB,\ QB\perp AB,\ PA=QB\text{ and }PQ\text{ intersects }AB\text{ at }M.
\displaystyle \text{To prove: }M\text{ is the midpoint of }AB\text{ and }PQ.
\displaystyle \text{Proof: In }\triangle APM\text{ and }\triangle BQM,
\displaystyle \angle PAM=\angle QBM=90^\circ
\displaystyle PA=QB\qquad\text{(given)}
\displaystyle \angle AMP=\angle BMQ\qquad\text{(vertically opposite angles)}
\displaystyle \therefore \triangle APM\cong\triangle BQM\qquad\text{(AAS congruency criterion).}
\displaystyle \therefore AM=BM\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Therefore, }M\text{ is the midpoint of }AB.
\displaystyle \text{Also, }PM=QM\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Therefore, }M\text{ is the midpoint of }PQ.
\displaystyle \text{Hence, }M\text{ is the midpoint of both }AB\text{ and }PQ.
\displaystyle \\

\displaystyle \textbf{Question 7: }AB\text{ is a line segment. }AX\text{ and }BY\text{ are two equal line segments}
\displaystyle \text{drawn on opposite sides of }AB\text{ such that }AX\parallel BY.\text{ If }AB\text{ and }XY
\displaystyle \text{intersect at }M,\text{ prove that:}
\displaystyle \text{(i) }\triangle AMX\cong\triangle BMY.

\displaystyle \text{(ii) }AB\text{ and }XY\text{ bisect each other at }M.
\displaystyle \text{Answer:} \displaystyle \text{Given: }AB\text{ is a line segment, }AX=BY,\ AX\parallel BY,\text{ and }AB\text{ and }XY
\displaystyle \text{intersect at }M.
\displaystyle \text{To prove: (i) }\triangle AMX\cong\triangle BMY.

\displaystyle \text{(ii) }AB\text{ and }XY\text{ bisect each other at }M.
\displaystyle \text{Proof: (i) In }\triangle AMX\text{ and }\triangle BMY,
\displaystyle AX=BY\qquad\text{(given)}
\displaystyle \angle MAX=\angle MBY\qquad\text{(alternate interior angles, as }AX\parallel BY\text{)}
\displaystyle \angle AXM=\angle BYM\qquad\text{(alternate interior angles, as }AX\parallel BY\text{)}
\displaystyle \therefore \triangle AMX\cong\triangle BMY\qquad\text{(ASA congruency criterion).}

\displaystyle \text{(ii) }AM=BM\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore M\text{ is the midpoint of }AB.
\displaystyle \text{Also, }XM=YM\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore M\text{ is the midpoint of }XY.
\displaystyle \therefore AB\text{ and }XY\text{ bisect each other at }M.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the given figure, the sides }BA\text{ and }CA\text{ of }\triangle ABC\text{ have}
\displaystyle \text{been produced to }D\text{ and }E\text{ respectively, such that }BA=AD\text{ and }CA=AE.
\displaystyle \text{Prove that }ED\parallel BC. \displaystyle \text{Answer:}
\displaystyle \text{Given: The sides }BA\text{ and }CA\text{ of }\triangle ABC\text{ are produced to }D\text{ and }E
\displaystyle \text{respectively, such that }BA=AD\text{ and }CA=AE.
\displaystyle \text{To prove: }ED\parallel BC.
\displaystyle \text{Proof: In }\triangle ABC\text{ and }\triangle ADE,
\displaystyle AB=AD\qquad\text{(given)}
\displaystyle AC=AE\qquad\text{(given)}
\displaystyle \angle BAC=\angle DAE\qquad\text{(vertically opposite angles)}
\displaystyle \therefore \triangle ABC\cong\triangle ADE\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore \angle ABC=\angle ADE\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{These are alternate interior angles formed by the transversal }BD.
\displaystyle \therefore ED\parallel BC.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the given figure, the line segments }AB\text{ and }CD\text{ intersect at}
\displaystyle \text{a point }M\text{ such that }AM=MD\text{ and }CM=MB.\text{ Prove that }AC=BD,
\displaystyle \text{but }AC\text{ may not be parallel to }BD. \displaystyle \text{Answer:}
\displaystyle \text{Given: }AB\text{ and }CD\text{ intersect at }M,\ AM=MD\text{ and }CM=MB.
\displaystyle \text{To prove: }AC=BD,\text{ but }AC\text{ may not be parallel to }BD.
\displaystyle \text{Proof: In }\triangle AMC\text{ and }\triangle DMB,
\displaystyle AM=MD\qquad\text{(given)}
\displaystyle CM=MB\qquad\text{(given)}
\displaystyle \angle AMC=\angle DMB\qquad\text{(vertically opposite angles)}
\displaystyle \therefore \triangle AMC\cong\triangle DMB\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore AC=BD\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Also, }\angle ACM=\angle DBM\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{These angles are not alternate interior angles, so they do not imply}
\displaystyle AC\parallel BD.
\displaystyle \therefore AC\text{ may not be parallel to }BD.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If two altitudes of a triangle are equal, prove that it is an}
\displaystyle \text{isosceles triangle.}
\displaystyle \text{Answer:} \displaystyle \text{Given: In }\triangle ABC,\ BD\perp AC,\ CE\perp AB\text{ and }BD=CE.
\displaystyle \text{To prove: }\triangle ABC\text{ is isosceles, i.e., }AB=AC.
\displaystyle \text{Proof: In right-angled }\triangle CBD\text{ and }\triangle BCE,
\displaystyle BD=CE\qquad\text{(given)}
\displaystyle BC=BC\qquad\text{(common hypotenuse)}
\displaystyle \therefore \triangle CBD\cong\triangle BCE\qquad\text{(RHS congruency criterion).}
\displaystyle \therefore \angle B=\angle C\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore AB=AC\qquad\text{(sides opposite equal angles are equal).}
\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the given figure, }\angle BAC=\angle CDB\text{ and }
\displaystyle \angle BCA=\angle CBD.\text{ Prove that }AB=CD. \displaystyle \text{Answer:}
\displaystyle \text{Given: }\angle BAC=\angle CDB\text{ and }\angle BCA=\angle CBD.
\displaystyle \text{To prove: }AB=CD.
\displaystyle \text{Proof: In }\triangle ABC\text{ and }\triangle DBC,
\displaystyle \angle BAC=\angle CDB\qquad\text{(given)}
\displaystyle \angle BCA=\angle CBD\qquad\text{(given)}
\displaystyle BC=BC\qquad\text{(common side)}
\displaystyle \therefore \triangle ABC\cong\triangle DBC\qquad\text{(AAS congruency criterion).}
\displaystyle \therefore AB=CD\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the given figure: }\angle ABD=\angle EBC,\ BD=BC\text{ and }
\displaystyle \angle ACB=\angle EDB.\text{ Prove that }AB=BE. \displaystyle \text{Answer:}
\displaystyle \text{Given: }\angle ABD=\angle EBC,\ BD=BC\text{ and }\angle ACB=\angle EDB.
\displaystyle \text{To prove: }AB=BE.
\displaystyle \text{Proof: }\angle ABD=\angle EBC\qquad\text{(given)}
\displaystyle \text{Adding }\angle ABE\text{ to both sides,}
\displaystyle \angle ABD+\angle ABE=\angle ABE+\angle EBC
\displaystyle \therefore \angle DBE=\angle CBA.
\displaystyle \text{Now in }\triangle ABC\text{ and }\triangle EBD,
\displaystyle BC=BD\qquad\text{(given)}
\displaystyle \angle ACB=\angle EDB\qquad\text{(given)}
\displaystyle \angle ABC=\angle DBE\qquad\text{(proved)}
\displaystyle \therefore \triangle ABC\cong\triangle EBD\qquad\text{(ASA congruency criterion).}
\displaystyle \therefore AB=BE\qquad\text{(C.P.C.T.C.).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the given figure, }AY\perp ZY\text{ and }BY\perp XY\text{ such that}
\displaystyle AY=ZY\text{ and }BY=XY.\text{ Prove that }AB=ZX. \displaystyle \text{Answer:}
\displaystyle \text{Given: }AY\perp ZY,\ BY\perp XY,\ AY=ZY\text{ and }BY=XY.
\displaystyle \text{To prove: }AB=ZX.
\displaystyle \text{Proof: }\angle XYB=90^\circ\text{ and }\angle AYZ=90^\circ.
\displaystyle \text{Adding }\angle AYX\text{ to both sides,}
\displaystyle \angle XYB+\angle AYX=\angle AYZ+\angle AYX
\displaystyle \therefore \angle AYB=\angle ZYX.
\displaystyle \text{Now in }\triangle AYB\text{ and }\triangle ZYX,
\displaystyle \angle AYB=\angle ZYX\qquad\text{(proved)}
\displaystyle AY=ZY\qquad\text{(given)}
\displaystyle BY=XY\qquad\text{(given)}
\displaystyle \therefore \triangle AYB\cong\triangle ZYX\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore AB=ZX\qquad\text{(C.P.C.T.C.).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the given figure, }ABCD\text{ is a square and }\triangle PAB\text{ is an}
\displaystyle \text{equilateral triangle.}

\displaystyle \text{(i) Prove that }\triangle APD\cong\triangle BPC.

\displaystyle \text{(ii) Show that }\angle DPC=30^\circ. \displaystyle \text{Answer:}
\displaystyle \text{Given: }ABCD\text{ is a square and }\triangle PAB\text{ is an equilateral triangle.}

\displaystyle \text{To prove: (i) }\triangle APD\cong\triangle BPC.

\displaystyle \text{(ii) }\angle DPC=30^\circ.
\displaystyle \text{Proof: Since }\triangle PAB\text{ is equilateral,}
\displaystyle PA=PB\text{ and }\angle PAB=\angle PBA=60^\circ.
\displaystyle \text{Also, since }ABCD\text{ is a square,}
\displaystyle AD=BC\text{ and }\angle DAB=\angle ABC=90^\circ.
\displaystyle \therefore \angle PAD=90^\circ+60^\circ=150^\circ
\displaystyle \text{and }\angle PBC=90^\circ+60^\circ=150^\circ.

\displaystyle \text{(i) In }\triangle APD\text{ and }\triangle BPC,
\displaystyle PA=PB
\displaystyle AD=BC
\displaystyle \angle PAD=\angle PBC=150^\circ
\displaystyle \therefore \triangle APD\cong\triangle BPC\qquad\text{(SAS congruency criterion).}

\displaystyle \text{(ii) From C.P.C.T.C., }PD=PC.
\displaystyle \therefore \triangle DPC\text{ is isosceles and }\angle PDC=\angle DCP.
\displaystyle \text{Also, }PA=AD=AB,
\displaystyle \therefore \angle ADP=\angle APD.
\displaystyle \text{Now in }\triangle APD,
\displaystyle \angle APD+\angle ADP+\angle PAD=180^\circ
\displaystyle \therefore 2\angle APD+150^\circ=180^\circ
\displaystyle \therefore 2\angle APD=30^\circ
\displaystyle \therefore \angle APD=15^\circ.
\displaystyle \text{Since }\angle APB=60^\circ,
\displaystyle \angle DPC=\angle APB-\angle APD-\angle BPC
\displaystyle =60^\circ-15^\circ-15^\circ=30^\circ.
\displaystyle \therefore \angle DPC=30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the given figure, in }\triangle ABC,\ \angle B=90^\circ.
\displaystyle \text{If }ABPQ\text{ and }ACRS\text{ are squares, prove that:}

\displaystyle \text{(i) }\triangle ACQ\cong\triangle ABS.

\displaystyle \text{(ii) }CQ=BS. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ \angle B=90^\circ,\text{ and }ABPQ\text{ and }ACRS
\displaystyle \text{are squares.}

\displaystyle \text{To prove: (i) }\triangle ACQ\cong\triangle ABS.

\displaystyle \text{(ii) }CQ=BS.
\displaystyle \text{Proof: }\angle CAQ=90^\circ+\angle BAC.
\displaystyle \angle BAS=90^\circ+\angle BAC.
\displaystyle \therefore \angle CAQ=\angle BAS.

\displaystyle \text{(i) In }\triangle ACQ\text{ and }\triangle ABS,
\displaystyle AQ=AB\qquad\text{(sides of square }ABPQ\text{)}
\displaystyle AC=AS\qquad\text{(sides of square }ACRS\text{)}
\displaystyle \angle CAQ=\angle BAS\qquad\text{(proved)}
\displaystyle \therefore \triangle ACQ\cong\triangle ABS\qquad\text{(SAS congruency criterion).}

\displaystyle \text{(ii) }\therefore CQ=BS\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Squares }ABPQ\text{ and }ADRS\text{ are drawn on the sides }AB\text{ and }
\displaystyle AD\text{ of a parallelogram }ABCD.\text{ Prove that:}

\displaystyle \text{(i) }\angle SAQ=\angle ABC.

\displaystyle \text{(ii) }SQ=AC.
\displaystyle \text{Answer:}

\displaystyle \text{Given: }ABPQ\text{ and }ADRS\text{ are squares drawn on the sides }AB
\displaystyle \text{and }AD\text{ of parallelogram }ABCD.
\displaystyle \text{Join }SQ\text{ and }AC.

\displaystyle \text{To prove: (i) }\angle SAQ=\angle ABC.

\displaystyle \text{(ii) }SQ=AC.
\displaystyle \text{Proof: (i) Reflex }\angle SAQ
\displaystyle =\angle SAD+\angle DAB+\angle BAQ
\displaystyle =90^\circ+\angle DAB+90^\circ
\displaystyle =180^\circ+\angle DAB.
\displaystyle \therefore \angle SAQ=360^\circ-(180^\circ+\angle DAB)
\displaystyle =180^\circ-\angle DAB.\qquad\cdots(1)
\displaystyle AB\parallel CD\qquad\text{(opposite sides of a parallelogram)}
\displaystyle \therefore \angle ABC+\angle DAB=180^\circ.
\displaystyle \therefore \angle ABC=180^\circ-\angle DAB.\qquad\cdots(2)
\displaystyle \text{From (1) and (2), }\angle SAQ=\angle ABC.

\displaystyle \text{(ii) In }\triangle SAQ\text{ and }\triangle ABC,
\displaystyle AS=AD\qquad\text{(sides of square }ADRS\text{)}
\displaystyle AD=BC\qquad\text{(opposite sides of a parallelogram)}
\displaystyle \therefore AS=BC.
\displaystyle AQ=AB\qquad\text{(sides of square }ABPQ\text{)}
\displaystyle \angle SAQ=\angle ABC\qquad\text{(proved)}
\displaystyle \therefore \triangle SAQ\cong\triangle ABC\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore SQ=AC\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the given figure, }ABCD\text{ is a parallelogram and }E\text{ is the}
\displaystyle \text{midpoint of }BC.\text{ }DE\text{ produced meets }AB\text{ produced at }L.\text{ Prove that:}

\displaystyle \text{(i) }AB=BL.

\displaystyle \text{(ii) }AL=2DC. \displaystyle \text{Answer:}
\displaystyle \text{Given: }ABCD\text{ is a parallelogram, }E\text{ is the midpoint of }BC,
\displaystyle \text{and }DE\text{ produced meets }AB\text{ produced at }L.

\displaystyle \text{To prove: (i) }AB=BL.

\displaystyle \text{(ii) }AL=2DC.
\displaystyle \text{Proof: Since }E\text{ is the midpoint of }BC,\ BE=EC.

\displaystyle \text{(i) In }\triangle EBL\text{ and }\triangle CED,
\displaystyle BE=EC
\displaystyle \angle BEL=\angle CED\qquad\text{(vertically opposite angles)}
\displaystyle \angle EBL=\angle ECD\qquad\text{(alternate interior angles)}
\displaystyle \therefore \triangle EBL\cong\triangle CED\qquad\text{(AAS congruency criterion).}
\displaystyle \therefore BL=CD\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Also, }AB=CD\qquad\text{(opposite sides of a parallelogram).}
\displaystyle \therefore AB=BL.

\displaystyle \text{(ii) }AL=AB+BL
\displaystyle =AB+AB=2AB
\displaystyle =2DC\qquad\text{(since }AB=DC\text{).}
\displaystyle \therefore AL=2DC.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Equilateral triangles }ABD\text{ and }ACE\text{ are drawn on the sides }AB
\displaystyle \text{and }AC\text{ of }\triangle ABC\text{ as shown in the figure. Prove that:}

\displaystyle \text{(i) }\angle DAC=\angle EAB.

\displaystyle \text{(ii) }DC=BE. \displaystyle \text{Answer:}
\displaystyle \text{Given: Equilateral triangles }ABD\text{ and }ACE\text{ are drawn on the sides }AB
\displaystyle \text{and }AC\text{ of }\triangle ABC.

\displaystyle \text{To prove: (i) }\angle DAC=\angle EAB.

\displaystyle \text{(ii) }DC=BE.
\displaystyle \text{Proof: }\angle DAC=\angle DAB+\angle BAC
\displaystyle =60^\circ+\angle BAC.\qquad\cdots(1)
\displaystyle \angle BAE=\angle CAE+\angle BAC
\displaystyle =60^\circ+\angle BAC.\qquad\cdots(2)
\displaystyle \text{From (1) and (2), }\angle DAC=\angle EAB.

\displaystyle \text{(ii) In }\triangle DAC\text{ and }\triangle BAE,
\displaystyle \angle DAC=\angle EAB\qquad\text{(proved)}
\displaystyle AD=AB\qquad\text{(sides of an equilateral triangle)}
\displaystyle AC=AE\qquad\text{(sides of an equilateral triangle)}
\displaystyle \therefore \triangle DAC\cong\triangle BAE\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore DC=BE\qquad\text{(C.P.C.T.C.).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In the given figure, }ABCD\text{ is a square and }P,\ Q,\ R\text{ are points}
\displaystyle \text{on }AB,\ BC\text{ and }CD\text{ respectively such that }AP=BQ=CR\text{ and}
\displaystyle \angle PQR=90^\circ.\text{ Prove that:}

\displaystyle \text{(i) }PB=QC\qquad\text{(ii) }PQ=QR\qquad\text{(iii) }\angle QPR=45^\circ. \displaystyle \text{Answer:}
\displaystyle \text{(i) Since }AB=BC\text{ (sides of a square) and }AP=BQ\text{ (given),}
\displaystyle PB=AB-AP=BC-BQ=QC.

\displaystyle \text{(ii) In }\triangle PBQ\text{ and }\triangle QCR,
\displaystyle \angle PQB=\angle QRC
\displaystyle \angle PQB=\angle QRC=90^\circ
\displaystyle PB=QC\qquad\text{(proved)}
\displaystyle BQ=CR\qquad\text{(given)}
\displaystyle \therefore \triangle PBQ\cong\triangle QCR\qquad\text{(AAS congruency criterion).}
\displaystyle \therefore PQ=QR\qquad\text{(C.P.C.T.C.).}

\displaystyle \text{(iii) In }\triangle PQR,
\displaystyle \angle PQR=90^\circ
\displaystyle PQ=QR
\displaystyle \therefore \angle QPR=\angle PRQ.
\displaystyle \angle QPR+\angle PRQ=90^\circ
\displaystyle \therefore 2\angle QPR=90^\circ
\displaystyle \therefore \angle QPR=45^\circ.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the given figure, }ABCD\text{ is a square, }EF\parallel BD\text{ and }R
\displaystyle \text{is the midpoint of }EF.\text{ Prove that:}

\displaystyle \text{(i) }BE=DF\qquad\text{(ii) }AR\text{ bisects }\angle BAD.

\displaystyle \text{(iii) }\text{If }AR\text{ is produced, it will pass through }C. \displaystyle \text{Answer:}
\displaystyle \text{Given: }ABCD\text{ is a square, }EF\parallel BD,\ R\text{ is the midpoint of }EF.
\displaystyle \text{Join }BD\text{ and }RC.

\displaystyle \text{(i) Since }\angle ABD=\angle ADB=45^\circ,
\displaystyle \text{and }BD\parallel EF,
\displaystyle \angle FEC=\angle ABD=45^\circ
\displaystyle \angle EFC=\angle ADB=45^\circ.
\displaystyle \therefore \angle FEC=\angle EFC.
\displaystyle \therefore EC=FC.
\displaystyle \text{Also, }BC=DC\qquad\text{(sides of a square).}
\displaystyle \therefore BC-EC=DC-FC
\displaystyle \therefore BE=DF.

\displaystyle \text{(ii) In }\triangle ABE\text{ and }\triangle ADF,
\displaystyle AB=AD\qquad\text{(sides of a square)}
\displaystyle \angle B=\angle D=90^\circ
\displaystyle BE=DF\qquad\text{(proved)}
\displaystyle \therefore \triangle ABE\cong\triangle ADF\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore AE=AF\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore \triangle AEF\text{ is isosceles.}
\displaystyle \text{Since }R\text{ is the midpoint of }EF,\ AR\text{ bisects }\angle EAF.
\displaystyle \text{Also, }BD\parallel EF,
\displaystyle \angle EAF=\angle BAD.
\displaystyle \therefore AR\text{ bisects }\angle BAD.

\displaystyle \text{(iii) The diagonal }BD\text{ bisects }\angle BAD\text{ of the square.}
\displaystyle \text{Hence }AR\text{ coincides with the diagonal }AC.
\displaystyle \therefore \text{if }AR\text{ is produced, it passes through }C.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{ABCD is a parallelogram in which }\angle A\text{ and }\angle C\text{ are}
\displaystyle \text{obtuse. Points }X\text{ and }Y\text{ are taken on diagonal }BD\text{ such that}
\displaystyle \angle AXD=\angle CYB=90^\circ.\text{ Prove that }XA=YC. \displaystyle \text{Answer:}
\displaystyle \text{Given: }ABCD\text{ is a parallelogram, }X\text{ and }Y\text{ lie on diagonal }BD,
\displaystyle \text{and }\angle AXD=\angle CYB=90^\circ.
\displaystyle \text{To prove: }XA=YC.
\displaystyle \text{Proof: In }\triangle ADX\text{ and }\triangle BCY,
\displaystyle AD=BC\qquad\text{(opposite sides of a parallelogram)}
\displaystyle \angle AXD=\angle CYB=90^\circ\qquad\text{(given)}
\displaystyle \angle ADX=\angle CBY\qquad\text{(alternate interior angles, as }AD\parallel BC\text{)}
\displaystyle \therefore \triangle ADX\cong\triangle BCY\qquad\text{(AAS congruency criterion).}
\displaystyle \therefore XA=YC\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{ABCD is a parallelogram. The sides }AB\text{ and }AD\text{ are produced}
\displaystyle \text{to }E\text{ and }F\text{ respectively, such that }AB=BE\text{ and }AD=DF.
\displaystyle \text{Prove that }\triangle BEC\cong\triangle DCF. \displaystyle \text{Answer:}
\displaystyle \text{Given: }ABCD\text{ is a parallelogram, }AB\text{ is produced to }E\text{ and }AD
\displaystyle \text{is produced to }F,\text{ such that }AB=BE\text{ and }AD=DF.
\displaystyle \text{To prove: }\triangle BEC\cong\triangle DCF.
\displaystyle \text{Proof: Since }ABCD\text{ is a parallelogram,}
\displaystyle AB=CD\text{ and }BC=AD.
\displaystyle \text{But }AB=BE\text{ and }AD=DF.
\displaystyle \therefore BE=CD\text{ and }BC=DF.
\displaystyle \text{Also, }AB\parallel CD\text{ and }AD\parallel BC.
\displaystyle \therefore \angle CBE=\angle FDC.
\displaystyle \text{Now in }\triangle BEC\text{ and }\triangle DCF,
\displaystyle BE=CD
\displaystyle BC=DF
\displaystyle \angle CBE=\angle FDC
\displaystyle \therefore \triangle BEC\cong\triangle DCF\qquad\text{(SAS congruency criterion).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The perpendicular bisectors of the sides of }\triangle ABC\text{ meet at }I.
\displaystyle \text{Prove that }IA=IB=IC. \displaystyle \text{Answer:}
\displaystyle \text{Given: The perpendicular bisectors of }AB,\ BC\text{ and }CA\text{ meet at }I.
\displaystyle \text{To prove: }IA=IB=IC.
\displaystyle \text{Proof: Let the perpendicular bisector of }AB\text{ meet }AB\text{ at }N.
\displaystyle \text{Then }AN=BN\text{ and }IN\perp AB.
\displaystyle \text{In }\triangle AIN\text{ and }\triangle BIN,
\displaystyle AN=BN\qquad\text{(since }N\text{ is the midpoint of }AB\text{)}
\displaystyle \angle ANI=\angle BNI=90^\circ
\displaystyle IN=IN\qquad\text{(common side)}
\displaystyle \therefore \triangle AIN\cong\triangle BIN\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore IA=IB\qquad\text{(C.P.C.T.C.).}\qquad\cdots(1)
\displaystyle \text{Let the perpendicular bisector of }AC\text{ meet }AC\text{ at }M.
\displaystyle \text{Then }AM=CM\text{ and }IM\perp AC.
\displaystyle \text{In }\triangle AIM\text{ and }\triangle CIM,
\displaystyle AM=CM\qquad\text{(since }M\text{ is the midpoint of }AC\text{)}
\displaystyle \angle AMI=\angle CMI=90^\circ
\displaystyle IM=IM\qquad\text{(common side)}
\displaystyle \therefore \triangle AIM\cong\triangle CIM\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore IA=IC\qquad\text{(C.P.C.T.C.).}\qquad\cdots(2)
\displaystyle \text{From (1) and (2), }IA=IB=IC.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Exercise 10(B)}

\displaystyle \textbf{Question 1: }\text{In }\triangle ABC,\ AB=AC\text{ and }\angle A=50^\circ.
\displaystyle \text{Find }\angle B\text{ and }\angle C.
\displaystyle \text{Answer:} \displaystyle \text{In }\triangle ABC,\ AB=AC\text{ and }\angle A=50^\circ.
\displaystyle \therefore \angle B=\angle C\qquad\text{(angles opposite to equal sides).}
\displaystyle \text{Also, }\angle A+\angle B+\angle C=180^\circ.
\displaystyle 50^\circ+\angle B+\angle B=180^\circ
\displaystyle 2\angle B=130^\circ
\displaystyle \therefore \angle B=65^\circ.
\displaystyle \therefore \angle C=65^\circ.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In }\triangle ABC,\ BC=AC\text{ and }\angle B=64^\circ.
\displaystyle \text{Find }\angle C.
\displaystyle \text{Answer:} \displaystyle \text{In }\triangle ABC,\ BC=AC\text{ and }\angle B=64^\circ.
\displaystyle \therefore \angle A=\angle B=64^\circ\qquad\text{(angles opposite to equal sides).}
\displaystyle \text{Also, }\angle A+\angle B+\angle C=180^\circ.
\displaystyle 64^\circ+64^\circ+\angle C=180^\circ
\displaystyle \therefore \angle C=180^\circ-128^\circ=52^\circ.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In each of the following figures, find the value of }x.

\displaystyle \text{(i)}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,\ AB=AC.
\displaystyle \therefore \angle B=\angle C\qquad\text{(angles opposite to equal sides).}
\displaystyle 40^\circ+\angle B+\angle B=180^\circ
\displaystyle 2\angle B=140^\circ
\displaystyle \therefore \angle B=\angle C=70^\circ.
\displaystyle \text{Since }\angle ACD\text{ is an exterior angle,}
\displaystyle x=\angle A+\angle B=40^\circ+70^\circ=110^\circ.

\displaystyle \text{(ii)}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ACD,\ AC=CD.
\displaystyle \therefore \angle CAD=\angle CDA=30^\circ.
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle CAD+\angle CDA+\angle ACD=180^\circ
\displaystyle 30^\circ+30^\circ+\angle ACD=180^\circ
\displaystyle \therefore \angle ACD=120^\circ.
\displaystyle \text{In }\triangle ABC,\ \angle ACD=\angle CAB+\angle ABC.
\displaystyle 120^\circ=65^\circ+x
\displaystyle \therefore x=55^\circ.

\displaystyle \text{(iii)}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,\ AB=AC.
\displaystyle \therefore \angle B=\angle C=55^\circ\qquad\text{(angles opposite to equal sides).}
\displaystyle \text{Since }B,\ C\text{ and }D\text{ are collinear,}
\displaystyle \angle ACD=180^\circ-55^\circ=125^\circ.
\displaystyle \text{In }\triangle ACD,
\displaystyle x+75^\circ+125^\circ=180^\circ
\displaystyle \therefore x=20^\circ.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In each of the following figures, find the value of }x.

\displaystyle \text{(i)}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABD,\ AD=BD.
\displaystyle \therefore \angle BAD=\angle ABD=50^\circ
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \text{In }\triangle ADC,\ AD=DC.
\displaystyle \therefore \angle DAC=\angle DCA=x
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle B+\angle BAC+\angle C=180^\circ
\displaystyle 50^\circ+(50^\circ+x)+x=180^\circ
\displaystyle 100^\circ+2x=180^\circ
\displaystyle 2x=80^\circ
\displaystyle \therefore x=40^\circ.

\displaystyle \text{(ii)}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ADB,\ AD=BD.
\displaystyle \therefore \angle DAB=\angle ABD=x
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \text{In }\triangle ADC,\ AD=DC.
\displaystyle \therefore \angle DAC=\angle ACD
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \angle ACE+\angle ACD=180^\circ\qquad\text{(linear pair)}
\displaystyle 124^\circ+\angle ACD=180^\circ
\displaystyle \therefore \angle ACD=56^\circ.
\displaystyle \therefore \angle DAC=56^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle B+\angle BAC+\angle BCA=180^\circ
\displaystyle x+(x+56^\circ)+56^\circ=180^\circ
\displaystyle 2x+112^\circ=180^\circ
\displaystyle 2x=68^\circ
\displaystyle \therefore x=34^\circ.

\displaystyle \text{(iii)}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABD,\ AD=BD.
\displaystyle \therefore \angle BAD=\angle ABD=x
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \text{In }\triangle ADC,\ AC=DC.
\displaystyle \therefore \angle ADC=\angle CAD
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \angle ADC+\angle CAD+\angle ACD=180^\circ
\displaystyle 2\angle ADC+64^\circ=180^\circ
\displaystyle 2\angle ADC=116^\circ
\displaystyle \therefore \angle ADC=58^\circ.
\displaystyle \text{Since }\angle ADC\text{ is an exterior angle of }\triangle ABD,
\displaystyle \angle ADC=\angle ABD+\angle BAD
\displaystyle 58^\circ=x+x
\displaystyle 2x=58^\circ
\displaystyle \therefore x=29^\circ.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the given figure, }BD\parallel CE,\ AC=BC,\ \angle ABD=20^\circ
\displaystyle \text{and }\angle ECF=70^\circ.\text{ Find }\angle GAC. \displaystyle \text{Answer:}
\displaystyle \text{Since }BD\parallel CE,
\displaystyle \angle DBC=\angle ECF=70^\circ\qquad\text{(corresponding angles).}
\displaystyle \angle ABC+\angle ABD=\angle DBC
\displaystyle \angle ABC+20^\circ=70^\circ
\displaystyle \therefore \angle ABC=50^\circ.
\displaystyle \text{In }\triangle ABC,\ AC=BC.
\displaystyle \therefore \angle BAC=\angle ABC=50^\circ
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \angle GAC+\angle BAC=180^\circ\qquad\text{(linear pair)}
\displaystyle \angle GAC+50^\circ=180^\circ
\displaystyle \therefore \angle GAC=130^\circ.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the given figure, }AB=AC,\ \angle A=50^\circ\text{ and}
\displaystyle \angle ACD=15^\circ.\text{ Show that }BC=CD. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB=AC,\ \angle A=50^\circ
\displaystyle \text{and }\angle ACD=15^\circ.
\displaystyle \text{To prove: }BC=CD.
\displaystyle \text{Proof: Since }AB=AC,
\displaystyle \angle ABC=\angle ACB
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \angle ABC+\angle ACB+\angle BAC=180^\circ
\displaystyle 2\angle ABC+50^\circ=180^\circ
\displaystyle 2\angle ABC=130^\circ
\displaystyle \therefore \angle ABC=65^\circ.
\displaystyle \text{Since }A,\ D\text{ and }B\text{ are collinear, }\angle BDC\text{ is an exterior angle}
\displaystyle \text{of }\triangle ADC.
\displaystyle \therefore \angle BDC=\angle DAC+\angle ACD
\displaystyle =50^\circ+15^\circ=65^\circ.
\displaystyle \therefore \angle ABC=\angle BDC=65^\circ.
\displaystyle \text{Hence, in }\triangle BCD,\text{ the sides opposite equal angles are equal.}
\displaystyle \therefore BC=CD.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In the given figure, }AB\parallel CD\text{ and }CA=CE.
\displaystyle \text{Find the values of }x,\ y\text{ and }z. \displaystyle \text{Answer:}
\displaystyle \text{Since }AB\parallel CD,
\displaystyle \angle BAD=\angle ADC\qquad\text{(alternate interior angles).}
\displaystyle \therefore x=36^\circ.
\displaystyle \text{In }\triangle ACE,\ AC=CE.
\displaystyle \therefore \angle CAE=\angle CEA=y
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \text{In }\triangle CDE,\ \angle CEA\text{ is an exterior angle.}
\displaystyle \therefore \angle CEA=\angle ECD+\angle EDC
\displaystyle y=32^\circ+36^\circ=68^\circ.
\displaystyle \text{In }\triangle ACE,
\displaystyle \angle ACE+\angle CEA+\angle CAE=180^\circ
\displaystyle z+68^\circ+68^\circ=180^\circ
\displaystyle z=180^\circ-136^\circ=44^\circ.
\displaystyle \therefore x=36^\circ,\ y=68^\circ\text{ and }z=44^\circ.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In an isosceles triangle, the exterior angles formed by producing the}
\displaystyle \text{equal sides are equal to each other. Prove this.}
\displaystyle \text{Answer:} \displaystyle \text{Given: In }\triangle ABC,\ AB=AC,\text{ and }BC\text{ is produced to }D\text{ and }E.
\displaystyle \text{To prove: }\angle ABD=\angle ACE.
\displaystyle \text{Proof: Since }AB=AC,
\displaystyle \angle ABC=\angle ACB\qquad\text{(angles opposite to equal sides).}\qquad\cdots(1)
\displaystyle \angle ABD+\angle ABC=180^\circ\qquad\text{(linear pair).}\qquad\cdots(2)
\displaystyle \angle ACE+\angle ACB=180^\circ\qquad\text{(linear pair).}\qquad\cdots(3)
\displaystyle \text{From (1), (2) and (3),}
\displaystyle \angle ABD=\angle ACE.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the given figure, side }CA\text{ of }\triangle ABC\text{ has been produced}
\displaystyle \text{to }E.\text{ If }AC=AD=BD,\ \angle ACD=46^\circ\text{ and }\angle BAE=x^\circ, \displaystyle \text{find the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ADC,\ AC=AD.
\displaystyle \therefore \angle ACD=\angle ADC=46^\circ.
\displaystyle \text{In }\triangle ADB,\ AD=BD.
\displaystyle \therefore \angle DAB=\angle DBA.
\displaystyle \text{Also, }\angle ADC=\angle DAB+\angle DBA\qquad\text{(exterior angle theorem).}
\displaystyle 46^\circ=2\angle DBA
\displaystyle \therefore \angle DBA=\angle DAB=23^\circ.
\displaystyle \text{Now in }\triangle ABC,
\displaystyle \angle BAE=\angle B+\angle C
\displaystyle =23^\circ+46^\circ=69^\circ.
\displaystyle \therefore x=69^\circ.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the given figure, }CA=CD=BD,\ \angle DBC=35^\circ\text{ and}
\displaystyle \angle DCA=x^\circ.\text{ Find the value of }x. \displaystyle \text{Answer:}
\displaystyle \text{In }\triangle BCD,\ BD=CD.
\displaystyle \therefore \angle DBC=\angle BCD=35^\circ.
\displaystyle \text{In }\triangle ACD,\ AC=CD.
\displaystyle \therefore \angle CAD=\angle CDA.
\displaystyle \text{Since }\angle CDA\text{ is an exterior angle of }\triangle BCD,
\displaystyle \angle CDA=\angle DBC+\angle BCD
\displaystyle =35^\circ+35^\circ=70^\circ.
\displaystyle \therefore \angle CAD=70^\circ.
\displaystyle \text{Now in }\triangle ACD,
\displaystyle \angle CDA+\angle CAD+\angle DCA=180^\circ
\displaystyle 70^\circ+70^\circ+x=180^\circ
\displaystyle \therefore x=40^\circ.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the given figure, }\triangle ABC\text{ is an equilateral triangle whose}
\displaystyle \text{base }BC\text{ is produced to }D\text{ such that }BC=CD.\text{ Calculate:}

\displaystyle \text{(i) }\angle ACD\qquad\text{(ii) }\angle ADC.
\displaystyle \text{Answer:}

\displaystyle \text{(i) Since }\triangle ABC\text{ is equilateral,}
\displaystyle \angle ABC=\angle BCA=\angle CAB=60^\circ.
\displaystyle \text{Hence, }\angle ACD=\angle ACB+\angle BCD
\displaystyle =60^\circ+60^\circ=120^\circ.

\displaystyle \text{(ii) In }\triangle ACD,\ AC=CD.
\displaystyle \therefore \angle CAD=\angle ADC.
\displaystyle \angle CAD+\angle ADC+\angle ACD=180^\circ
\displaystyle 2\angle ADC+120^\circ=180^\circ
\displaystyle \therefore \angle ADC=30^\circ.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the given figure, }AB=AD,\ CB=CD,\ \angle A=42^\circ\text{ and}
\displaystyle \angle C=108^\circ.\text{ Find }\angle ABC. \displaystyle \text{Answer:}
\displaystyle \text{Join }BD.\displaystyle \text{In }\triangle ABD,\ AB=AD.
\displaystyle \therefore \angle ABD=\angle ADB.
\displaystyle \angle A+\angle ABD+\angle ADB=180^\circ
\displaystyle 42^\circ+2\angle ABD=180^\circ
\displaystyle \therefore \angle ABD=69^\circ.\qquad\cdots(1)
\displaystyle \text{In }\triangle BCD,\ CB=CD.
\displaystyle \therefore \angle CBD=\angle BDC.
\displaystyle \angle CBD+\angle BDC+\angle C=180^\circ
\displaystyle 2\angle CBD+108^\circ=180^\circ
\displaystyle \therefore \angle CBD=36^\circ.\qquad\cdots(2)
\displaystyle \text{From (1) and (2),}
\displaystyle \angle ABC=\angle ABD+\angle DBC
\displaystyle =69^\circ+36^\circ=105^\circ.
\displaystyle \therefore \angle ABC=105^\circ.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the given figure, side }BA\text{ of }\triangle ABC\text{ has been produced}
\displaystyle \text{to }D\text{ such that }CD=CA,\text{ and side }CB\text{ has been produced to }E.
\displaystyle \text{If }\angle BAC=106^\circ\text{ and }\angle ABE=128^\circ,\text{ find }\angle BCD. \displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ACD,\ AC=CD.
\displaystyle \therefore \angle CAD=\angle CDA
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \angle BAC+\angle CAD=180^\circ\qquad\text{(linear pair)}
\displaystyle 106^\circ+\angle CAD=180^\circ
\displaystyle \therefore \angle CAD=74^\circ.
\displaystyle \therefore \angle CDA=74^\circ.
\displaystyle \text{Since }B,\ A\text{ and }D\text{ are collinear,}
\displaystyle \angle CDB=\angle CDA=74^\circ.
\displaystyle \text{In }\triangle BCD,\ \angle DBE\text{ is an exterior angle.}
\displaystyle \therefore \angle DBE=\angle BCD+\angle CDB
\displaystyle 128^\circ=\angle BCD+74^\circ
\displaystyle \therefore \angle BCD=54^\circ.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the given figure, }AB=BC\text{ and }AC=CD.\text{ Show that}
\displaystyle \angle BAD:\angle ADB=3:1. \displaystyle \text{Answer:}
\displaystyle \text{Given: }AB=BC\text{ and }AC=CD.
\displaystyle \text{To prove: }\angle BAD:\angle ADB=3:1.
\displaystyle \text{Proof: In }\triangle ABC,\ AB=BC.
\displaystyle \therefore \angle BAC=\angle BCA
\displaystyle \qquad\text{(angles opposite to equal sides).}\qquad\cdots(1)
\displaystyle \text{In }\triangle ACD,\ AC=CD.
\displaystyle \therefore \angle CAD=\angle CDA
\displaystyle \qquad\text{(angles opposite to equal sides).}\qquad\cdots(2)
\displaystyle \text{Since }\angle ACB\text{ is an exterior angle of }\triangle ACD,
\displaystyle \angle ACB=\angle CAD+\angle CDA
\displaystyle =2\angle CDA.\qquad\cdots(3)
\displaystyle \text{From (1) and (3),}
\displaystyle \angle BAC=2\angle CDA.
\displaystyle \text{Now, }\angle BAD=\angle BAC+\angle CAD
\displaystyle =2\angle CDA+\angle CDA
\displaystyle =3\angle CDA.
\displaystyle \text{Since }B,\ C\text{ and }D\text{ are collinear,}
\displaystyle \angle ADB=\angle CDA.
\displaystyle \therefore \angle BAD=3\angle ADB.
\displaystyle \therefore \angle BAD:\angle ADB=3:1.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Show that the perpendiculars drawn from the extremities of the base}
\displaystyle \text{of an isosceles triangle to the opposite sides are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB=AC,\ BD\perp AC\text{ and }CE\perp AB.
\displaystyle \text{To prove: }BD=CE.
\displaystyle \text{Proof: Since }AB=AC,

\displaystyle \angle ABC=\angle ACB
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \text{In }\triangle BDC\text{ and }\triangle CEB,
\displaystyle \angle BDC=\angle CEB=90^\circ
\displaystyle \angle BCD=\angle CBE\qquad\text{(since }\angle ACB=\angle ABC\text{)}
\displaystyle BC=BC\qquad\text{(common side)}
\displaystyle \therefore \triangle BDC\cong\triangle CEB\qquad\text{(AAS congruency criterion).}
\displaystyle \therefore BD=CE\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In }\triangle ABC,\ AB=AC.\text{ If the bisectors of }\angle B\text{ and }\angle C
\displaystyle \text{meet }AC\text{ and }AB\text{ at }D\text{ and }E\text{ respectively, show that:}
\displaystyle \text{(i) }\triangle DBC\cong\triangle ECB.

\displaystyle \text{(ii) }BD=CE.

\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB=AC,\text{ and }BD\text{ and }CE\text{ bisect }\angle B
\displaystyle \text{and }\angle C\text{ respectively.}

\displaystyle \text{To prove: (i) }\triangle DBC\cong\triangle ECB.

\displaystyle \text{(ii) }BD=CE.
\displaystyle \text{Proof: Since }AB=AC,
\displaystyle \angle ABC=\angle ACB
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \text{Since }BD\text{ and }CE\text{ bisect }\angle B\text{ and }\angle C\text{ respectively,}
\displaystyle \angle DBC=\angle ECB.

\displaystyle \text{(i) In }\triangle DBC\text{ and }\triangle ECB,
\displaystyle \angle DBC=\angle ECB
\displaystyle \angle DCB=\angle EBC\qquad\text{(since }\angle ACB=\angle ABC\text{)}
\displaystyle BC=CB\qquad\text{(common side)}
\displaystyle \therefore \triangle DBC\cong\triangle ECB\qquad\text{(ASA congruency criterion).}

\displaystyle \text{(ii) }\therefore BD=CE\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In an isosceles triangle, prove that the altitude from the vertex}
\displaystyle \text{bisects the base.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB=AC\text{ and }AD\perp BC.
\displaystyle \text{To prove: }AD\text{ bisects }BC,\text{ i.e., }BD=DC.
\displaystyle \text{Proof: In right-angled }\triangle ABD\text{ and }\triangle ACD,
\displaystyle AB=AC\qquad\text{(given hypotenuses)}
\displaystyle AD=AD\qquad\text{(common side)}
\displaystyle \therefore \triangle ABD\cong\triangle ACD\qquad\text{(RHS congruency criterion).}
\displaystyle \therefore BD=DC\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore AD\text{ bisects }BC.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If the altitude from one vertex of a triangle bisects the opposite side,}
\displaystyle \text{prove that the triangle is isosceles.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AD\perp BC\text{ and }BD=DC.
\displaystyle \text{To prove: }\triangle ABC\text{ is isosceles, i.e., }AB=AC.
\displaystyle \text{Proof: In }\triangle ABD\text{ and }\triangle ACD,\displaystyle AD=AD\qquad\text{(common side)}
\displaystyle \angle ADB=\angle ADC=90^\circ
\displaystyle BD=DC\qquad\text{(given)}
\displaystyle \therefore \triangle ABD\cong\triangle ACD\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore AB=AC\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In the given figure, }AD=AE\text{ and }\angle BAD=\angle CAE.
\displaystyle \text{Prove that }AB=AC. \displaystyle \text{Answer:}
\displaystyle \text{Given: }AD=AE\text{ and }\angle BAD=\angle CAE.
\displaystyle \text{To prove: }AB=AC.
\displaystyle \text{Proof: In }\triangle ADE,
\displaystyle AD=AE\qquad\text{(given)}
\displaystyle \therefore \angle ADE=\angle AED
\displaystyle \qquad\text{(angles opposite to equal sides).}
\displaystyle \angle ADE+\angle ADB=180^\circ\qquad\text{(linear pair)}
\displaystyle \angle AED+\angle AEC=180^\circ\qquad\text{(linear pair)}
\displaystyle \therefore \angle ADB=\angle AEC.
\displaystyle \text{Now in }\triangle ABD\text{ and }\triangle ACE,
\displaystyle AD=AE\qquad\text{(given)}
\displaystyle \angle ADB=\angle AEC\qquad\text{(proved)}
\displaystyle \angle BAD=\angle CAE\qquad\text{(given)}
\displaystyle \therefore \triangle ABD\cong\triangle ACE\qquad\text{(ASA congruency criterion).}
\displaystyle \therefore AB=AC\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the given figure, }AB=AC,\ D\text{ is the midpoint of }BC,
\displaystyle DP\perp BA\text{ and }DQ\perp CA.\text{ Prove that:}

\displaystyle \text{(i) }DP=DQ\qquad\text{(ii) }AP=AQ\qquad\text{(iii) }AD\text{ bisects }\angle A. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB=AC,\ D\text{ is the midpoint of }BC,
\displaystyle DP\perp BA\text{ and }DQ\perp CA.

\displaystyle \text{To prove: (i) }DP=DQ\qquad\text{(ii) }AP=AQ\qquad\text{(iii) }AD\text{ bisects }\angle A.
\displaystyle \text{Proof: Since }AB=AC,
\displaystyle \angle B=\angle C\qquad\text{(angles opposite to equal sides).}
\displaystyle \angle BPD=\angle CQD=90^\circ
\displaystyle BD=DC\qquad\text{(}D\text{ is the midpoint of }BC\text{).}
\displaystyle \text{In }\triangle BPD\text{ and }\triangle CQD,
\displaystyle \angle BPD=\angle CQD,\ \angle PBD=\angle QCD,\ BD=DC
\displaystyle \therefore \triangle BPD\cong\triangle CQD\qquad\text{(AAS congruency criterion).}
\displaystyle \therefore DP=DQ\qquad\text{(C.P.C.T.C.).}

\displaystyle \text{(ii) Since }AB=AC\text{ and }BP=CQ,
\displaystyle AP=AB-BP=AC-CQ=AQ.

\displaystyle \text{(iii) In }\triangle APD\text{ and }\triangle AQD,
\displaystyle AD=AD\qquad\text{(common side)}
\displaystyle DP=DQ\qquad\text{(proved)}
\displaystyle AP=AQ\qquad\text{(proved)}
\displaystyle \therefore \triangle APD\cong\triangle AQD\qquad\text{(SSS congruency criterion).}
\displaystyle \therefore \angle PAD=\angle QAD\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore AD\text{ bisects }\angle A.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In the given figure, }AB=AC.\text{ If }BO\text{ and }CO\text{ are the}
\displaystyle \text{bisectors of }\angle B\text{ and }\angle C\text{ respectively, meeting at }O,\text{ and }BC
\displaystyle \text{is produced to }D,\text{ prove that }\angle BOC=\angle ACD. \displaystyle \text{Answer:}
\displaystyle \text{Given: }AB=AC,\ BO\text{ and }CO\text{ are the bisectors of }\angle B
\displaystyle \text{and }\angle C,\text{ and }BC\text{ is produced to }D.
\displaystyle \text{To prove: }\angle BOC=\angle ACD.
\displaystyle \text{Proof: Since }AB=AC,
\displaystyle \angle ABC=\angle ACB.
\displaystyle \text{As }BO\text{ and }CO\text{ bisect }\angle B\text{ and }\angle C,
\displaystyle \angle OBC=\angle OCB.
\displaystyle \text{Hence, in }\triangle BOC,
\displaystyle \angle BOC=180^\circ-(\angle OBC+\angle OCB)
\displaystyle =180^\circ-2\angle OCB
\displaystyle =180^\circ-\angle ACB.\qquad\cdots(1)
\displaystyle \angle ACD+\angle ACB=180^\circ\qquad\text{(linear pair).}\qquad\cdots(2)
\displaystyle \text{From (1) and (2),}
\displaystyle \angle BOC=\angle ACD.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Prove that the bisectors of the base angles of an isosceles triangle}
\displaystyle \text{are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB=AC,\ BD\text{ and }CE\text{ are the bisectors}
\displaystyle \text{of }\angle B\text{ and }\angle C\text{ respectively.}
\displaystyle \text{To prove: }BD=CE.
\displaystyle \text{Proof: Since }AB=AC,\displaystyle \angle B=\angle C\qquad\text{(angles opposite to equal sides).}
\displaystyle \text{Therefore, }\angle DBC=\angle ECB.
\displaystyle \text{Also, }\angle DCB=\angle EBC.
\displaystyle \text{In }\triangle BCD\text{ and }\triangle CBE,
\displaystyle BC=CB\qquad\text{(common side)}
\displaystyle \angle DBC=\angle ECB
\displaystyle \angle DCB=\angle EBC
\displaystyle \therefore \triangle BCD\cong\triangle CBE\qquad\text{(ASA congruency criterion).}
\displaystyle \therefore BD=CE\qquad\text{(C.P.C.T.C.).}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In the given figure, }AB=AC.\text{ Side }BA\text{ has been produced to }D.
\displaystyle \text{If }AE\text{ is the bisector of }\angle CAD,\text{ prove that }AE\parallel BC. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB=AC,\ BA\text{ is produced to }D,
\displaystyle \text{and }AE\text{ bisects }\angle CAD.
\displaystyle \text{To prove: }AE\parallel BC.
\displaystyle \text{Proof: Since }AB=AC,
\displaystyle \angle B=\angle C\qquad\text{(angles opposite to equal sides).}
\displaystyle \text{Also, }\angle CAD=\angle B+\angle C
\displaystyle =2\angle C\qquad\text{(exterior angle theorem).}
\displaystyle \therefore \angle C=\frac{1}{2}\angle CAD.\qquad\cdots(1)
\displaystyle \text{Since }AE\text{ bisects }\angle CAD,
\displaystyle \angle EAC=\frac{1}{2}\angle CAD.\qquad\cdots(2)
\displaystyle \text{From (1) and (2), }\angle EAC=\angle C.
\displaystyle \text{These are alternate interior angles.}
\displaystyle \therefore AE\parallel BC.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{In the given figure, }AD\text{ is the internal bisector of }\angle A\text{ and }
\displaystyle CE\parallel DA.\text{ If }CE\text{ meets }BA\text{ produced at }E,\text{ prove that }\triangle CAE \displaystyle \text{is isosceles.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AD\text{ bisects }\angle A,\ CE\parallel DA,
\displaystyle \text{and }BA\text{ is produced to }E.
\displaystyle \text{To prove: }\triangle CAE\text{ is isosceles.}
\displaystyle \text{Proof: In }\triangle ACE,
\displaystyle \angle CAE=\angle E+\angle ACE\qquad\text{(exterior angle theorem).}
\displaystyle \text{Since }AD\parallel CE,
\displaystyle \angle DAC=\angle ACE\qquad\text{(alternate interior angles)}
\displaystyle \text{and }\angle BAD=\angle AEC\qquad\text{(corresponding angles).}
\displaystyle \text{But }\angle BAD=\angle DAC\qquad\text{(}AD\text{ bisects }\angle A\text{).}
\displaystyle \therefore \angle ACE=\angle AEC.
\displaystyle \therefore AE=AC\qquad\text{(sides opposite equal angles are equal).}
\displaystyle \therefore \triangle ACE\text{ is isosceles.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{In the adjoining figure, }AB=AC.\text{ If }DB\perp BC\text{ and }EC\perp BC,
\displaystyle \text{prove that: (i) }BD=CE\qquad\text{(ii) }AD=AE. \displaystyle \text{Answer:}
\displaystyle \text{Given: }AB=AC,\ DB\perp BC\text{ and }EC\perp BC.
\displaystyle \text{To prove: (i) }BD=CE\qquad\text{(ii) }AD=AE.
\displaystyle \text{Proof: Since }AB=AC,
\displaystyle \angle ABC=\angle ACB\qquad\text{(angles opposite to equal sides).}
\displaystyle \text{Also, }DB\perp BC\text{ and }EC\perp BC,
\displaystyle DB\parallel EC.
\displaystyle \therefore \angle ABD=\angle ACE\qquad\text{(alternate interior angles).}
\displaystyle \text{Also, }\angle BAD=\angle CAE\qquad\text{(vertically opposite angles).}
\displaystyle \text{Now in }\triangle ABD\text{ and }\triangle ACE,
\displaystyle AB=AC
\displaystyle \angle BAD=\angle CAE
\displaystyle \angle ABD=\angle ACE
\displaystyle \therefore \triangle ABD\cong\triangle ACE\qquad\text{(ASA congruency criterion).}
\displaystyle \therefore BD=CE\qquad\text{and}\qquad AD=AE\qquad\text{(C.P.C.T.C.).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{In the given figure, }\triangle ABC\text{ is an equilateral triangle and }BC
\displaystyle \text{is produced to }D\text{ such that }BC=CD.\text{ Prove that }AD\perp AB. \displaystyle \text{Answer:}
\displaystyle \text{Given: }\triangle ABC\text{ is equilateral and }BC\text{ is produced to }D
\displaystyle \text{such that }BC=CD.
\displaystyle \text{To prove: }AD\perp AB.
\displaystyle \text{Proof: Since }\triangle ABC\text{ is equilateral,}
\displaystyle AB=BC=CA\text{ and }\angle A=\angle B=\angle C=60^\circ.
\displaystyle \text{In }\triangle ACD,\ AC=CD.
\displaystyle \therefore \angle CAD=\angle CDA.
\displaystyle \text{Also, }\angle ACB=\angle CAD+\angle CDA.
\displaystyle 60^\circ=2\angle CAD
\displaystyle \therefore \angle CAD=30^\circ.
\displaystyle \therefore \angle BAD=\angle BAC+\angle CAD
\displaystyle =60^\circ+30^\circ=90^\circ.
\displaystyle \therefore AD\perp AB.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{In the given figure, }AB=AC,\ AC\text{ is the bisector of }\angle A,
\displaystyle AD=CD\text{ and }\angle ABC=75^\circ.\text{ Find }x\text{ and }y. \displaystyle \text{Answer:}
\displaystyle \text{Since }AB=AC,
\displaystyle \angle ABC=\angle ACB=75^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle x+75^\circ+75^\circ=180^\circ
\displaystyle \therefore x=30^\circ.
\displaystyle \text{In }\triangle ACD,\ AD=CD.
\displaystyle \therefore \angle DAC=\angle ACD.
\displaystyle \text{Since }AC\text{ bisects }\angle A,
\displaystyle \angle DAC=\angle CAB=x=30^\circ.
\displaystyle \therefore \angle ACD=30^\circ.
\displaystyle \text{Now in }\triangle ACD,
\displaystyle \angle DAC+\angle ACD+\angle ADC=180^\circ
\displaystyle 30^\circ+30^\circ+y=180^\circ
\displaystyle \therefore y=120^\circ.
\displaystyle \therefore x=30^\circ,\ y=120^\circ.
\displaystyle \\

\displaystyle \textbf{Exercise 10(C)}

\displaystyle \textbf{Question 1: }\text{In }\triangle PQR,\ \angle P=50^\circ\text{ and }\angle R=70^\circ.
\displaystyle \text{Name (i) the shortest side (ii) the longest side.}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle PQR,\displaystyle \angle Q=180^\circ-(50^\circ+70^\circ)=60^\circ.
\displaystyle \text{The side opposite the smaller angle is the shortest, and the side opposite}
\displaystyle \text{the greater angle is the longest.}
\displaystyle \text{(i) }\angle P=50^\circ\text{ is the smallest angle.}
\displaystyle \therefore QR\text{ is the shortest side.}
\displaystyle \text{(ii) }\angle R=70^\circ\text{ is the greatest angle.}
\displaystyle \therefore PQ\text{ is the longest side.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In }\triangle LMN,\ \angle M=90^\circ.\text{ Name the longest side.}
\displaystyle \text{Answer:} \displaystyle \text{In }\triangle LMN,\ \angle M=90^\circ.
\displaystyle \text{The side opposite the right angle is the hypotenuse.}
\displaystyle \text{The hypotenuse is the longest side of a right-angled triangle.}
\displaystyle \therefore LN\text{ is the longest side.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the given figure, side }AB\text{ of }\triangle ABC\text{ is produced to }D
\displaystyle \text{such that }BD=BC.\text{ If }\angle A=60^\circ\text{ and }\angle B=50^\circ,\text{ prove that:}

\displaystyle \text{(i) }AD>CD\qquad\text{(ii) }AD>AC. \displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle C=180^\circ-(60^\circ+50^\circ)=70^\circ.
\displaystyle \text{In }\triangle BCD,\ BD=BC.
\displaystyle \therefore \angle BCD=\angle BDC.
\displaystyle \text{Since }\angle CBA\text{ is an exterior angle of }\triangle BCD,
\displaystyle \angle CBA=\angle BCD+\angle BDC.
\displaystyle 50^\circ=2\angle BDC
\displaystyle \therefore \angle BDC=\angle BCD=25^\circ.
\displaystyle \therefore \angle ACD=70^\circ+25^\circ=95^\circ.
\displaystyle \text{(i) }\angle ACD>\angle CAB
\displaystyle \therefore AD>CD\qquad\text{(side opposite the greater angle is longer).}
\displaystyle \text{(ii) }\angle ACD>\angle ADC
\displaystyle \therefore AD>AC\qquad\text{(side opposite the greater angle is longer).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In a right-angled triangle, prove that the hypotenuse is the}
\displaystyle \text{longest side.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: A right-angled triangle }ABC\text{ in which }\angle B=90^\circ.
\displaystyle \text{To prove: }AC\text{ is the longest side.}
\displaystyle \text{Proof: In }\triangle ABC,\displaystyle \angle B=90^\circ.
\displaystyle \therefore \angle A+\angle C=90^\circ.
\displaystyle \therefore \angle A<90^\circ\text{ and }\angle C<90^\circ.
\displaystyle \therefore \angle B>\angle A\text{ and }\angle B>\angle C.
\displaystyle \text{The side opposite the greater angle is longer.}
\displaystyle \therefore AC>AB\text{ and }AC>BC.
\displaystyle \therefore AC\text{ is the longest side.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the given figure, }AB>AC.\text{ If }BO\text{ and }CO\text{ are the}
\displaystyle \text{bisectors of }\angle B\text{ and }\angle C\text{ respectively, prove that }BO>CO. \displaystyle \text{Answer:}
\displaystyle \text{Since }AB>AC,
\displaystyle \angle C>\angle B\qquad\text{(angle opposite the longer side is greater).}
\displaystyle \text{As }BO\text{ and }CO\text{ bisect }\angle B\text{ and }\angle C,
\displaystyle \angle OCB>\angle OBC.
\displaystyle \text{In }\triangle BOC,
\displaystyle \text{the side opposite the greater angle is longer.}
\displaystyle \therefore BO>CO.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the given figure, sides }AB\text{ and }AC\text{ of }\triangle ABC\text{ have}
\displaystyle \text{been produced to }D\text{ and }E\text{ respectively. If }\angle CBD=x^\circ \displaystyle \text{and }\angle BCE=y^\circ\text{ such that }x>y,\text{ show that }AB>AC.
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle ABC+x=180^\circ\qquad\text{(linear pair).}\qquad\cdots(1)
\displaystyle \angle ACB+y=180^\circ\qquad\text{(linear pair).}\qquad\cdots(2)
\displaystyle \text{From (1) and (2),}
\displaystyle \angle ABC=180^\circ-x,\qquad \angle ACB=180^\circ-y.
\displaystyle \text{Since }x>y,
\displaystyle 180^\circ-x<180^\circ-y.
\displaystyle \therefore \angle ABC<\angle ACB.
\displaystyle \text{The side opposite the greater angle is longer.}
\displaystyle \therefore AB>AC.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In the adjoining figure, }AB=AC.\text{ If }CB\text{ is produced to }D,
\displaystyle \text{show that }AD>AB. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB=AC,\ CB\text{ is produced to }D
\displaystyle \text{and }AD\text{ is joined.}
\displaystyle \text{To prove: }AD>AB.
\displaystyle \text{Proof: In }\triangle ABD,
\displaystyle \angle ADB>\angle ACB\qquad\text{(exterior angle theorem).}
\displaystyle \text{Since }AB=AC,
\displaystyle \angle B=\angle C\qquad\text{(angles opposite to equal sides).}
\displaystyle \therefore \angle ADB>\angle C=\angle DCA.
\displaystyle \text{In }\triangle ACD,
\displaystyle \text{the side opposite the greater angle is longer.}
\displaystyle \therefore AD>AC.
\displaystyle \text{But }AB=AC.
\displaystyle \therefore AD>AB.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the adjoining figure, }AB>AC\text{ and }D\text{ is any point on }BC.
\displaystyle \text{Show that }AB>AD. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AB>AC,\ D\text{ is any point on }BC
\displaystyle \text{and }AD\text{ is joined.}
\displaystyle \text{To prove: }AB>AD.
\displaystyle \text{Proof: In }\triangle ACD,
\displaystyle \angle ADB>\angle ACD\qquad\text{(exterior angle theorem).}
\displaystyle \text{Since }AB>AC,
\displaystyle \angle C>\angle B\qquad\text{(angle opposite the longer side is greater).}
\displaystyle \therefore \angle ADB>\angle B.
\displaystyle \text{In }\triangle ABD,
\displaystyle \text{the side opposite the greater angle is longer.}
\displaystyle \therefore AB>AD.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the adjoining figure, }AC>AB\text{ and }AD\text{ is the bisector of }
\displaystyle \angle A.\text{ Show that }\angle ADC>\angle ADB. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ AC>AB\text{ and }AD\text{ bisects }\angle A,
\displaystyle \text{meeting }BC\text{ at }D.
\displaystyle \text{To prove: }\angle ADC>\angle ADB.
\displaystyle \text{Proof: Since }AC>AB,
\displaystyle \angle B>\angle C\qquad\text{(angle opposite the longer side is greater).}
\displaystyle \text{Adding }\angle BAD\text{ to both sides,}
\displaystyle \angle B+\angle BAD>\angle C+\angle BAD.
\displaystyle \text{Since }AD\text{ bisects }\angle A,
\displaystyle \angle BAD=\angle CAD.
\displaystyle \therefore \angle B+\angle BAD>\angle C+\angle CAD.
\displaystyle \text{But }\angle B+\angle BAD=\angle ADC
\displaystyle \text{and }\angle C+\angle CAD=\angle ADB
\displaystyle \qquad\text{(exterior angle theorem).}
\displaystyle \therefore \angle ADC>\angle ADB.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the adjoining figure, in }\triangle ABC,\ O\text{ is any point in its}
\displaystyle \text{interior. Show that }OB+OC<AB+AC. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ O\text{ is a point in the interior of }\triangle ABC.
\displaystyle \text{Join }OB,\ OC\text{ and produce }BO\text{ to meet }AC\text{ at }D.
\displaystyle \text{To prove: }OB+OC<AB+AC.
\displaystyle \text{Proof: In }\triangle ABD,
\displaystyle AB+AD>BD\qquad\text{(sum of two sides of a triangle is greater than the third).}
\displaystyle \text{Since }BD=BO+OD,
\displaystyle AB+AD>BO+OD.\qquad\cdots(1)
\displaystyle \text{In }\triangle OCD,
\displaystyle OD+OC>AC\qquad\text{(sum of two sides of a triangle is greater than the third).}\qquad\cdots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle AB+AD+OD+OC>BO+OD+AC.
\displaystyle \text{Since }A,\ O,\ D\text{ are collinear, }AD=AO+OD.
\displaystyle \text{Cancelling }AD\text{ and }OD\text{ appropriately gives}
\displaystyle AB+AC>BO+OC.
\displaystyle \therefore OB+OC<AB+AC.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In }\triangle ABC,\ D\text{ is any point on }BC.\text{ Prove that}
\displaystyle AB+BC+AC>2AD. \displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ D\text{ is a point on }BC\text{ and }AD\text{ is joined.}
\displaystyle \text{To prove: }AB+BC+AC>2AD.
\displaystyle \text{Proof: In }\triangle ABD,
\displaystyle AB+BD>AD
\displaystyle \qquad\text{(sum of two sides of a triangle is greater than the third side).}\qquad\cdots(1)
\displaystyle \text{Similarly, in }\triangle ACD,
\displaystyle AC+CD>AD
\displaystyle \qquad\text{(sum of two sides of a triangle is greater than the third side).}\qquad\cdots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle AB+BD+AC+CD>2AD.
\displaystyle \text{Since }BD+CD=BC,
\displaystyle \therefore AB+BC+AC>2AD.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the adjoining figure, }O\text{ is the centre of a circle, }XY\text{ is a}
\displaystyle \text{diameter and }XZ\text{ is a chord. Prove that }XY>XZ. \displaystyle \text{Answer:}
\displaystyle \text{Given: }O\text{ is the centre of the circle, }XY\text{ is a diameter and }XZ
\displaystyle \text{is a chord.}
\displaystyle \text{To prove: }XY>XZ.
\displaystyle \text{Construction: Join }OZ.
\displaystyle \text{Proof: In }\triangle OXZ,
\displaystyle OX+OZ>XZ
\displaystyle \qquad\text{(sum of two sides of a triangle is greater than the third side).}
\displaystyle \text{Since }OX=OZ=OY\qquad\text{(radii of the same circle),}
\displaystyle OX+OY>XZ.
\displaystyle \text{But }OX+OY=XY.
\displaystyle \therefore XY>XZ.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the adjoining figure, }PL\perp QR,\ LQ=LS\text{ and }LR>LQ.
\displaystyle \text{Show that }PR>PQ. \displaystyle \text{Answer:}
\displaystyle \text{Given: }PL\perp QR,\ LQ=LS\text{ and }LR>LQ.
\displaystyle \text{To prove: }PR>PQ.
\displaystyle \text{Construction: Join }PS.
\displaystyle \text{Proof: In }\triangle PQL\text{ and }\triangle PSL,
\displaystyle PL=PL\qquad\text{(common side)}
\displaystyle \angle PLQ=\angle PLS=90^\circ
\displaystyle LQ=LS\qquad\text{(given)}
\displaystyle \therefore \triangle PQL\cong\triangle PSL\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore PQ=PS\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore \angle PQS=\angle QSP
\displaystyle \qquad\text{(angles opposite to equal sides in }\triangle PQS\text{).}
\displaystyle \text{In }\triangle PSR,\ \angle QSP\text{ is an exterior angle.}
\displaystyle \therefore \angle QSP>\angle PRS.
\displaystyle \therefore \angle PQS>\angle PRS.
\displaystyle \text{Since }Q,\ S\text{ and }R\text{ are collinear, }\angle PQS=\angle PQR
\displaystyle \text{and }\angle PRS=\angle PRQ.
\displaystyle \therefore \angle PQR>\angle PRQ.
\displaystyle \text{Hence, in }\triangle PQR,\text{ the side opposite the greater angle is longer.}
\displaystyle \therefore PR>PQ.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the adjoining quadrilateral }ABCD,\ AB\text{ is the longest side and }CD
\displaystyle \text{is the shortest side. Prove that:}

\displaystyle \text{(i) }\angle C>\angle A\qquad\text{(ii) }\angle D>\angle B. \displaystyle \text{Answer:}
\displaystyle \text{Given: In quadrilateral }ABCD,\ AB\text{ is the longest side and }CD
\displaystyle \text{is the shortest side.}

\displaystyle \text{To prove: (i) }\angle C>\angle A.

\displaystyle \text{(ii) }\angle D>\angle B.
\displaystyle \text{Proof: (i) Join }AC.\displaystyle \text{In }\triangle ABC,\ AB>BC.
\displaystyle \therefore \angle ACB>\angle CAB
\displaystyle \qquad\text{(angle opposite the longer side is greater).}\qquad\cdots(1)
\displaystyle \text{In }\triangle ACD,\ AD>CD.
\displaystyle \therefore \angle DCA>\angle CAD
\displaystyle \qquad\text{(angle opposite the longer side is greater).}\qquad\cdots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle \angle ACB+\angle DCA>\angle CAB+\angle CAD.
\displaystyle \therefore \angle C>\angle A.

\displaystyle \text{(ii) Join }BD.
\displaystyle \text{In }\triangle ABD,\ AB>AD.
\displaystyle \therefore \angle ADB>\angle ABD
\displaystyle \qquad\text{(angle opposite the longer side is greater).}\qquad\cdots(3)
\displaystyle \text{In }\triangle BCD,\ BC>CD.
\displaystyle \therefore \angle BDC>\angle DBC
\displaystyle \qquad\text{(angle opposite the longer side is greater).}\qquad\cdots(4)
\displaystyle \text{Adding (3) and (4),}
\displaystyle \angle ADB+\angle BDC>\angle ABD+\angle DBC.
\displaystyle \therefore \angle D>\angle B.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Can you construct a }\triangle ABC\text{ in which }AB=5\text{ cm},
\displaystyle BC=4\text{ cm and }AC=9\text{ cm? Give a reason.}
\displaystyle \text{Answer:}
\displaystyle AB=5\text{ cm},\ BC=4\text{ cm and }AC=9\text{ cm}.
\displaystyle AB+BC=5+4=9\text{ cm}.
\displaystyle \therefore AB+BC=AC.
\displaystyle \text{However, the sum of any two sides of a triangle must be greater than the}
\displaystyle \text{third side.}
\displaystyle \therefore \text{it is not possible to construct such a triangle.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoining figure, }\triangle ABC\text{ is equilateral and }D\text{ is}
\displaystyle \text{any point on }AC.\text{ Prove that:}

\displaystyle \text{(i) }BD>AD\qquad\text{(ii) }BD>DC. \displaystyle \text{Answer:}
\displaystyle \text{Given: }\triangle ABC\text{ is equilateral, }D\text{ is a point on }AC
\displaystyle \text{and }BD\text{ is joined.}

\displaystyle \text{To prove: (i) }BD>AD\qquad\text{(ii) }BD>DC.
\displaystyle \text{Proof: Since }\triangle ABC\text{ is equilateral,}
\displaystyle AB=BC=CA\text{ and }\angle A=\angle B=\angle C=60^\circ.

\displaystyle \text{(i) Since }BD\text{ lies inside }\angle ABC,
\displaystyle \angle ABD<\angle ABC=60^\circ.
\displaystyle \text{Also, since }D\text{ lies on }AC,
\displaystyle \angle BAD=\angle BAC=60^\circ.
\displaystyle \therefore \angle BAD>\angle ABD.
\displaystyle \text{In }\triangle ABD,\text{ the side opposite the greater angle is longer.}
\displaystyle \therefore BD>AD.

\displaystyle \text{(ii) Since }BD\text{ lies inside }\angle ABC,
\displaystyle \angle DBC<\angle ABC=60^\circ.
\displaystyle \text{Also, since }D\text{ lies on }AC,
\displaystyle \angle BCD=\angle BCA=60^\circ.
\displaystyle \therefore \angle BCD>\angle DBC.
\displaystyle \text{In }\triangle BCD,\text{ the side opposite the greater angle is longer.}
\displaystyle \therefore BD>DC.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }O\text{ is any point inside }\triangle ABC,\text{ prove that}
\displaystyle \angle BOC>\angle A.
\displaystyle \text{Answer:}
\displaystyle \text{Given: In }\triangle ABC,\ O\text{ is any point inside the triangle.}
\displaystyle \text{Join }AO,\ BO\text{ and }CO.\displaystyle \text{Produce }AO\text{ to meet }BC\text{ at }D.
\displaystyle \text{To prove: }\angle BOC>\angle A.
\displaystyle \text{Proof: In }\triangle AOB,
\displaystyle \angle BOD>\angle ABO+\angle BAO
\displaystyle \qquad\text{(exterior angle theorem).}\qquad\cdots(1)
\displaystyle \text{In }\triangle AOC,
\displaystyle \angle COD>\angle ACO+\angle CAO
\displaystyle \qquad\text{(exterior angle theorem).}\qquad\cdots(2)
\displaystyle \text{Adding (1) and (2),}
\displaystyle \angle BOD+\angle DOC>\angle ABO+\angle ACO+\angle BAO+\angle CAO.
\displaystyle \therefore \angle BOC>\angle B+\angle C+\angle A=180^\circ-\angle A.
\displaystyle \text{Since }\angle B+\angle C=180^\circ-\angle A,
\displaystyle \therefore \angle BOC>\angle A.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the given figure, }AD=AB\text{ and }AE\text{ bisects }\angle A.\text{ Prove that:}

\displaystyle \text{(i) }BE=ED\qquad\text{(ii) }\angle ABD>\angle BCA. \displaystyle \text{Answer:}

\displaystyle \text{(i) In }\triangle ABE\text{ and }\triangle ADE,
\displaystyle AE=AE\qquad\text{(common side)}
\displaystyle \angle BAE=\angle EAD\qquad\text{(}AE\text{ bisects }\angle A\text{)}
\displaystyle AB=AD\qquad\text{(given)}
\displaystyle \therefore \triangle ABE\cong\triangle ADE\qquad\text{(SAS congruency criterion).}
\displaystyle \therefore BE=ED\qquad\text{(C.P.C.T.C.).}

\displaystyle \text{(ii) Also, }\angle ABE=\angle ADE\qquad\text{(C.P.C.T.C.).}
\displaystyle \therefore \angle ABD=\angle ADB.
\displaystyle \text{In }\triangle BCD,\ \angle ADB\text{ is an exterior angle.}
\displaystyle \therefore \angle ADB>\angle BCA.
\displaystyle \therefore \angle ABD>\angle BCA.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The sides }AB\text{ and }AC\text{ of }\triangle ABC\text{ are produced to }D
\displaystyle \text{and }E\text{ respectively. The bisectors of }\angle CBD\text{ and }\angle BCE
\displaystyle \text{meet at }O.\text{ If }AB>AC,\text{ prove that }OC>OB. \displaystyle \text{Answer:}
\displaystyle \text{Given: }AB>AC,\text{ and }BO,\ CO\text{ are the bisectors of }\angle CBD
\displaystyle \text{and }\angle BCE\text{ respectively.}
\displaystyle \text{To prove: }OC>OB.
\displaystyle \text{Proof: Since }AB>AC,
\displaystyle \angle C>\angle B\qquad\text{(angle opposite the longer side is greater).}
\displaystyle \angle ACB+\angle BCE=180^\circ
\displaystyle \angle ABC+\angle CBD=180^\circ
\displaystyle \therefore \angle BCE<\angle CBD.
\displaystyle \text{Since }CO\text{ and }BO\text{ bisect these angles,}
\displaystyle \angle OCB<\angle OBC.
\displaystyle \text{In }\triangle BOC,
\displaystyle \text{the side opposite the greater angle is longer.}
\displaystyle \therefore OC>OB.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In }\triangle ABC,\ AB=7.5\text{ cm},\ BC=6.2\text{ cm and }AC=5.4\text{ cm.}
\displaystyle \text{Name: (i) the least angle (ii) the greatest angle of the triangle.} \displaystyle \text{Answer:}
\displaystyle AC=5.4\text{ cm is the shortest side.}
\displaystyle \therefore \angle B\text{ is the least angle.}
\displaystyle AB=7.5\text{ cm is the longest side.}
\displaystyle \therefore \angle C\text{ is the greatest angle.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In }\triangle ABC,\text{ the angles are }\angle B=60^\circ,\ \angle C=40^\circ.
\displaystyle \text{Arrange }AB,\ BD\text{ and }DC\text{ in ascending order of their lengths.} \displaystyle \text{Answer:}
\displaystyle \angle A=180^\circ-(60^\circ+40^\circ)=80^\circ.
\displaystyle \text{Since }AD\text{ bisects }\angle A,
\displaystyle \angle BAD=\angle CAD=40^\circ.
\displaystyle \text{In }\triangle ABD,
\displaystyle \angle ADB=180^\circ-(60^\circ+40^\circ)=80^\circ.
\displaystyle \therefore AB>BD>AD.
\displaystyle \text{In }\triangle ACD,
\displaystyle AD=DC\qquad\text{(}\angle CAD=\angle ACD=40^\circ\text{).}
\displaystyle \therefore DC=AD.
\displaystyle \text{Hence }DC<BD<AB.
\displaystyle \\

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