\displaystyle \textbf{Question 1: }\text{In the adjoining figure, }X\text{ and }Y\text{ are two points on the equal sides }AB
\displaystyle \text{and }AC\text{ of }\triangle ABC\text{ such that }AX=AY.\text{ Prove that }XC=YB.

\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle AYB\text{ and }\triangle AXC.
\displaystyle AY=AX\qquad\text{(given)}
\displaystyle AB=AC\qquad\text{(given)}
\displaystyle \angle YAB=\angle XAC\qquad\text{(each is the angle between }AB\text{ and }AC\text{)}
\displaystyle \therefore \triangle AYB\cong\triangle AXC\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore YB=XC\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }D\text{ is the midpoint of the hypotenuse }AC\text{ of a right-angled}
\displaystyle \triangle ABC,\text{ prove that }BD=\frac{1}{2}AC.
\displaystyle \text{Answer:} \displaystyle \text{Produce }BD\text{ to }E\text{ such that }BD=DE,\text{ and join }CE.
\displaystyle \text{Consider }\triangle ADB\text{ and }\triangle CDE.
\displaystyle AD=DC\qquad\text{(since }D\text{ is the midpoint of }AC\text{)}
\displaystyle BD=DE\qquad\text{(by construction)}
\displaystyle \angle ADB=\angle CDE\qquad\text{(vertically opposite angles)}
\displaystyle \therefore \triangle ADB\cong\triangle CDE\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore AB=CE\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \text{Also, }\angle ABD=\angle CED\qquad\text{(corresponding angles)}
\displaystyle \text{Since }B,D,E\text{ are collinear, }\angle ABD=\angle ABE.
\displaystyle \therefore \angle ABE=\angle CEB
\displaystyle \therefore AB\parallel CE\qquad\text{(alternate interior angles are equal)}
\displaystyle \therefore \angle ABC+\angle BCE=180^\circ
\displaystyle \text{But }\angle ABC=90^\circ.
\displaystyle \therefore \angle BCE=90^\circ
\displaystyle \text{Now, in }\triangle ABC\text{ and }\triangle ECB,
\displaystyle AB=EC\qquad\text{(proved above)}
\displaystyle BC=CB\qquad\text{(common side)}
\displaystyle \angle ABC=\angle ECB=90^\circ
\displaystyle \therefore \triangle ABC\cong\triangle ECB\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore AC=EB\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \text{Since }BD=DE,\quad BE=BD+DE=2BD.
\displaystyle \therefore BD=\frac{1}{2}BE=\frac{1}{2}AC.
\displaystyle \therefore BD=\frac{1}{2}AC.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In a quadrilateral }ACBD,\text{ }AC=AD\text{ and }AB\text{ bisects }\angle A.
\displaystyle \text{Show that }\triangle ABC\cong\triangle ABD.\text{ What can you say about }BC\text{ and }BD?
\displaystyle \text{Answer:} Slide5\displaystyle \text{Consider }\triangle ABC\text{ and }\triangle ABD.
\displaystyle AC=AD\qquad\text{(given)}
\displaystyle \angle CAB=\angle DAB\qquad\text{(since }AB\text{ bisects }\angle A\text{)}
\displaystyle AB=AB\qquad\text{(common side)}
\displaystyle \therefore \triangle ABC\cong\triangle ABD\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore BC=BD\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Prove that }\triangle ABC\text{ is isosceles if any of the following holds:}
\displaystyle \text{(i) Altitude }AD\text{ bisects }BC.\qquad\text{(ii) Median }AD\text{ is perpendicular to }BC.
\displaystyle \text{Answer:} \displaystyle \text{(i) }AD\text{ is an altitude and bisects }BC.
\displaystyle \text{Consider }\triangle ADB\text{ and }\triangle ADC.
\displaystyle BD=DC\qquad\text{(since }AD\text{ bisects }BC\text{)}
\displaystyle \angle ADB=\angle ADC=90^\circ\qquad\text{(since }AD\perp BC\text{)}
\displaystyle AD=AD\qquad\text{(common side)}
\displaystyle \therefore \triangle ADB\cong\triangle ADC\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore AB=AC\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}

\displaystyle \text{(ii) }AD\text{ is a median and }AD\perp BC.
\displaystyle \text{Consider }\triangle ADB\text{ and }\triangle ADC.
\displaystyle BD=DC\qquad\text{(since }AD\text{ is a median)}
\displaystyle \angle ADB=\angle ADC=90^\circ\qquad\text{(since }AD\perp BC\text{)}
\displaystyle AD=AD\qquad\text{(common side)}
\displaystyle \therefore \triangle ADB\cong\triangle ADC\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore AB=AC\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the adjoining figure, }PS=QR\text{ and }\angle SPQ=\angle RQP.
\displaystyle \text{Prove that }\triangle PQS\cong\triangle QPR,\ PR=QS\text{ and }\angle QPR=\angle PQS. 2018-07-13_8-10-40.jpg\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle PQS\text{ and }\triangle QPR.
\displaystyle PS=QR\qquad\text{(given)}
\displaystyle \angle SPQ=\angle RQP\qquad\text{(given)}
\displaystyle PQ=QP\qquad\text{(common side)}
\displaystyle \therefore \triangle PQS\cong\triangle QPR\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore PR=QS\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \therefore \angle QPR=\angle PQS\qquad\text{(corresponding angles of congruent triangles)}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the adjoining figure, }AC=AE,\ AB=AD\text{ and }\angle BAD=\angle EAC. Slide7\displaystyle \text{Prove that }BC=DE.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AB=AD,\ AC=AE\text{ and }\angle BAD=\angle EAC.
\displaystyle \text{Adding }\angle DAC\text{ to both sides of }\angle BAD=\angle EAC,
\displaystyle \angle BAD+\angle DAC=\angle EAC+\angle DAC
\displaystyle \therefore \angle BAC=\angle DAE
\displaystyle \text{Consider }\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{(proved above)}
\displaystyle \therefore \triangle ABC\cong\triangle ADE\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore BC=DE\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In }\triangle PQR,\text{ }PQ=QR,\text{ and }L,\ M\text{ and }N\text{ are the midpoints}
\displaystyle \text{of the sides }PQ,\ QR\text{ and }RP\text{ respectively. Prove that }LN=MN.
\displaystyle \text{Answer:} \displaystyle \text{Since }L\text{ is the midpoint of }PQ,\quad PL=LQ=\frac{1}{2}PQ.
\displaystyle \text{Since }M\text{ is the midpoint of }QR,\quad QM=MR=\frac{1}{2}QR.
\displaystyle \text{But }PQ=QR\qquad\text{(given)}
\displaystyle \therefore PL=MR
\displaystyle \text{Also, }PN=NR\qquad\text{(since }N\text{ is the midpoint of }PR\text{)}
\displaystyle \text{Since }PQ=QR,\quad \angle QPR=\angle PRQ
\displaystyle \qquad\text{(angles opposite equal sides of a triangle are equal)}
\displaystyle \text{As }L\text{ lies on }PQ,\ M\text{ lies on }QR\text{ and }N\text{ lies on }PR,
\displaystyle \angle LPN=\angle MRN
\displaystyle \text{Consider }\triangle PLN\text{ and }\triangle MRN.
\displaystyle PL=MR\qquad\text{(proved above)}
\displaystyle PN=RN\qquad\text{(proved above)}
\displaystyle \angle LPN=\angle MRN\qquad\text{(proved above)}
\displaystyle \therefore \triangle PLN\cong\triangle MRN\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore LN=MN\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the adjoining figure, }PQRS\text{ is a square and }SRT\text{ is an equilateral}
\displaystyle \text{triangle. Prove that (i) }PT=QT\qquad\text{(ii) }\angle TQR=15^\circ. Slide8.JPG\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle PST\text{ and }\triangle QRT.
\displaystyle ST=RT\qquad\text{(sides of the equilateral triangle }SRT\text{)}
\displaystyle SP=RQ\qquad\text{(sides of the square }PQRS\text{)}
\displaystyle \angle PST=\angle PSR+\angle RST=90^\circ+60^\circ=150^\circ
\displaystyle \angle QRT=\angle QRS+\angle SRT=90^\circ+60^\circ=150^\circ
\displaystyle \therefore \angle PST=\angle QRT
\displaystyle \therefore \triangle PST\cong\triangle QRT\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore PT=QT\qquad\text{(corresponding sides of congruent triangles)}

\displaystyle \text{(ii) In }\triangle TQR,
\displaystyle RT=SR\qquad\text{(sides of the equilateral triangle }SRT\text{)}
\displaystyle SR=RQ\qquad\text{(sides of the square }PQRS\text{)}
\displaystyle \therefore RT=RQ
\displaystyle \therefore \angle QTR=\angle TQR\qquad\text{(angles opposite equal sides are equal)}
\displaystyle \text{Let }\angle QTR=\angle TQR=x.
\displaystyle \angle TRQ=150^\circ
\displaystyle \therefore x+x+150^\circ=180^\circ
\displaystyle 2x=30^\circ
\displaystyle x=15^\circ
\displaystyle \therefore \angle TQR=15^\circ.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove that the medians of an equilateral triangle are equal.}
\displaystyle \text{Answer:} \displaystyle \text{Let }AD,\ BE\text{ and }CF\text{ be the medians of the equilateral }\triangle ABC,
\displaystyle \text{where }D,\ E\text{ and }F\text{ are the midpoints of }BC,\ CA\text{ and }AB\text{ respectively.}
\displaystyle \text{Since }\triangle ABC\text{ is equilateral,}
\displaystyle AB=BC=CA\quad\text{and}\quad\angle A=\angle B=\angle C=60^\circ.
\displaystyle \text{Consider }\triangle ABE\text{ and }\triangle BCF.
\displaystyle AB=BC\qquad\text{(sides of an equilateral triangle)}
\displaystyle AE=BF\qquad\text{(halves of the equal sides }AC\text{ and }AB\text{)}
\displaystyle \angle BAE=\angle CBF=60^\circ
\displaystyle \therefore \triangle ABE\cong\triangle BCF\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore BE=CF\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \text{Now, consider }\triangle BCF\text{ and }\triangle CAD.
\displaystyle BC=CA\qquad\text{(sides of an equilateral triangle)}
\displaystyle BF=CD\qquad\text{(halves of the equal sides }AB\text{ and }BC\text{)}
\displaystyle \angle CBF=\angle ACD=60^\circ
\displaystyle \therefore \triangle BCF\cong\triangle CAD\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore CF=AD\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \text{Hence, }AD=BE=CF.
\displaystyle \therefore \text{The medians of an equilateral triangle are equal.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{ }AB\text{ is a line segment. }P\text{ and }Q\text{ are points on opposite sides of }AB
\displaystyle \text{such that each of them is equidistant from }A\text{ and }B.\text{ Show that }PQ\text{ is the}
\displaystyle \text{perpendicular bisector of }AB.
\displaystyle \text{Answer:} \displaystyle \text{Let }PQ\text{ intersect }AB\text{ at }M.
\displaystyle \text{Consider }\triangle APQ\text{ and }\triangle BPQ.
\displaystyle AP=BP\qquad\text{(since }P\text{ is equidistant from }A\text{ and }B\text{)}
\displaystyle AQ=BQ\qquad\text{(since }Q\text{ is equidistant from }A\text{ and }B\text{)}
\displaystyle PQ=PQ\qquad\text{(common side)}
\displaystyle \therefore \triangle APQ\cong\triangle BPQ\qquad\text{(by the SSS congruence criterion)}
\displaystyle \therefore \angle APQ=\angle BPQ\qquad\text{(corresponding angles of congruent triangles)}
\displaystyle \text{Since }M\text{ lies on }PQ,\quad \angle APM=\angle BPM.
\displaystyle \text{Now, consider }\triangle APM\text{ and }\triangle BPM.
\displaystyle AP=BP\qquad\text{(given)}
\displaystyle PM=PM\qquad\text{(common side)}
\displaystyle \angle APM=\angle BPM\qquad\text{(proved above)}
\displaystyle \therefore \triangle APM\cong\triangle BPM\qquad\text{(by the SAS congruence criterion)}
\displaystyle \therefore AM=BM\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \text{Also, }\angle AMP=\angle BMP\qquad\text{(corresponding angles)}
\displaystyle \text{But }\angle AMP+\angle BMP=180^\circ\qquad\text{(linear pair)}
\displaystyle \therefore \angle AMP=\angle BMP=90^\circ
\displaystyle \therefore PQ\perp AB.
\displaystyle \text{Also, }AM=BM,\text{ so }M\text{ is the midpoint of }AB.
\displaystyle \therefore PQ\text{ is the perpendicular bisector of }AB.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the adjoining figure, }AB=AC\text{ and }DB=DC.\text{ Find the ratio} Slide12\displaystyle \angle ABD:\angle ACD.
\displaystyle \text{Answer:}
\displaystyle \text{Since }AB=AC,\quad \angle ABC=\angle ACB
\displaystyle \qquad\text{(angles opposite equal sides of an isosceles triangle are equal).}
\displaystyle \text{Also, since }DB=DC,\quad \angle DBC=\angle DCB
\displaystyle \qquad\text{(angles opposite equal sides of an isosceles triangle are equal).}
\displaystyle \angle ABC=\angle ABD+\angle DBC
\displaystyle \angle ACB=\angle ACD+\angle DCB
\displaystyle \text{Since }\angle ABC=\angle ACB\text{ and }\angle DBC=\angle DCB,
\displaystyle \angle ABD+\angle DBC=\angle ACD+\angle DCB
\displaystyle \therefore \angle ABD=\angle ACD
\displaystyle \therefore \angle ABD:\angle ACD=1:1.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the adjoining figure, }\angle BCD=\angle ADC\text{ and }\angle ACB=\angle BDA.
\displaystyle \text{Prove that }AD=BC\text{ and }\angle CAD=\angle CBD. 2018-07-14_19-22-31\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle BCD=\angle ADC\text{ and }\angle ACB=\angle BDA.
\displaystyle \text{Adding the equal angles,}
\displaystyle \angle BCD+\angle ACB=\angle ADC+\angle BDA
\displaystyle \therefore \angle ACD=\angle BDC
\displaystyle \text{Now, consider }\triangle ACD\text{ and }\triangle BDC.
\displaystyle \angle ADC=\angle BCD\qquad\text{(given)}
\displaystyle \angle ACD=\angle BDC\qquad\text{(proved above)}
\displaystyle CD=DC\qquad\text{(common side)}
\displaystyle \therefore \triangle ACD\cong\triangle BDC\qquad\text{(by the ASA congruence criterion)}
\displaystyle \therefore AD=BC\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \therefore \angle CAD=\angle CBD\qquad\text{(corresponding angles of congruent triangles)}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the adjoining figure, }AC=BC,\ \angle DCA=\angle ECB\text{ and}
\displaystyle \angle DBC=\angle EAC.\text{ Prove that }\triangle DBC\cong\triangle EAC,\text{ and hence} 2018-07-14_19-23-21\displaystyle DC=EC\text{ and }BD=AE.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle DCA=\angle ECB.
\displaystyle \text{Adding }\angle DCE\text{ to both sides,}
\displaystyle \angle DCA+\angle DCE=\angle ECB+\angle DCE
\displaystyle \therefore \angle ECA=\angle DCB
\displaystyle \text{Now, consider }\triangle DBC\text{ and }\triangle EAC.
\displaystyle \angle DCB=\angle ECA\qquad\text{(proved above)}
\displaystyle BC=AC\qquad\text{(given)}
\displaystyle \angle DBC=\angle EAC\qquad\text{(given)}
\displaystyle \therefore \triangle DBC\cong\triangle EAC\qquad\text{(by the ASA congruence criterion)}
\displaystyle \therefore DC=EC\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \therefore BD=AE\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the adjoining figure, it is given that }CE=ED,\ \angle AOC=2\angle BCD
\displaystyle \text{and }\angle BOD=2\angle ADC.\text{ Prove that }\triangle CBE\cong\triangle DAE. 2018-07-19_7-12-48.jpg\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle CED,\quad CE=DE\qquad\text{(given)}
\displaystyle \therefore \angle ECD=\angle EDC
\displaystyle \qquad\text{(angles opposite equal sides of an isosceles triangle are equal)}
\displaystyle \angle AOC=\angle BOD\qquad\text{(vertically opposite angles)}
\displaystyle \text{But }\angle AOC=2\angle BCD\text{ and }\angle BOD=2\angle ADC.
\displaystyle \therefore 2\angle BCD=2\angle ADC
\displaystyle \therefore \angle BCD=\angle ADC
\displaystyle \text{Subtracting }\angle BCD=\angle ADC\text{ from }\angle ECD=\angle EDC,
\displaystyle \angle ECD-\angle BCD=\angle EDC-\angle ADC
\displaystyle \therefore \angle ECB=\angle EDA
\displaystyle \text{Now, consider }\triangle CBE\text{ and }\triangle DAE.
\displaystyle \angle ECB=\angle EDA\qquad\text{(proved above)}
\displaystyle CE=DE\qquad\text{(given)}
\displaystyle \angle CEB=\angle DEA
\displaystyle \qquad\text{(since }A,C,E\text{ and }B,D,E\text{ are collinear)}
\displaystyle \therefore \triangle CBE\cong\triangle DAE\qquad\text{(by the ASA congruence criterion)}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{ }BD\text{ and }CE\text{ are the bisectors of }\angle ABC\text{ and }\angle ACB
\displaystyle \text{respectively, of an isosceles }\triangle ABC\text{ in which }AB=AC.\text{ Prove that }BD=CE.
\displaystyle \text{Answer:} \displaystyle \text{Since }AB=AC,\quad \angle ABC=\angle ACB
\displaystyle \qquad\text{(angles opposite equal sides of an isosceles triangle are equal).}
\displaystyle \text{Since }BD\text{ bisects }\angle ABC,
\displaystyle \angle ABD=\frac{1}{2}\angle ABC
\displaystyle \text{Since }CE\text{ bisects }\angle ACB,
\displaystyle \angle ACE=\frac{1}{2}\angle ACB
\displaystyle \therefore \angle ABD=\angle ACE
\displaystyle \text{Now, consider }\triangle ABD\text{ and }\triangle ACE.
\displaystyle AB=AC\qquad\text{(given)}
\displaystyle \angle ABD=\angle ACE\qquad\text{(proved above)}
\displaystyle \angle BAD=\angle CAE\qquad\text{(each is equal to }\angle BAC\text{)}
\displaystyle \therefore \triangle ABD\cong\triangle ACE\qquad\text{(by the ASA congruence criterion)}
\displaystyle \therefore BD=CE\qquad\text{(corresponding sides of congruent triangles)}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoining figure, }\triangle ABC\text{ is an obtuse triangle, obtuse at}
\displaystyle \angle B.\text{ If }AD\perp BC,\text{ prove that }AC^2=AB^2+BC^2+2BC\times BD. 2018-10-13_15-21-36\displaystyle \text{Answer:}
\displaystyle \text{Since }AD\perp BC\text{ and }D,B,C\text{ are collinear, }\angle ADB=\angle ADC=90^\circ.
\displaystyle \text{In right-angled }\triangle ADB,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AD^2+BD^2\qquad\ldots\text{(i)}
\displaystyle \text{In right-angled }\triangle ADC,\text{ by the Pythagoras theorem,}
\displaystyle AC^2=AD^2+DC^2
\displaystyle \text{Since }D,B,C\text{ are collinear and }B\text{ lies between }D\text{ and }C,
\displaystyle DC=DB+BC
\displaystyle \therefore AC^2=AD^2+(DB+BC)^2
\displaystyle =AD^2+DB^2+BC^2+2DB\times BC
\displaystyle =(AD^2+DB^2)+BC^2+2DB\times BC
\displaystyle \therefore AC^2=AB^2+BC^2+2BC\times BD\qquad\text{[Using (i)]}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, }\angle B\text{ of }\triangle ABC\text{ is acute and}
\displaystyle AD\perp BC.\text{ Prove that }AC^2=AB^2+BC^2-2BC\times BD. 2018-10-13_15-21-48\displaystyle \text{Answer:}
\displaystyle \text{Since }AD\perp BC\text{ and }B,D,C\text{ are collinear, }\angle ADB=\angle ADC=90^\circ.
\displaystyle \text{In right-angled }\triangle ADB,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AD^2+BD^2\qquad\ldots\text{(i)}
\displaystyle \text{In right-angled }\triangle ADC,\text{ by the Pythagoras theorem,}
\displaystyle AC^2=AD^2+DC^2
\displaystyle \text{Since }D\text{ lies between }B\text{ and }C,
\displaystyle DC=BC-BD
\displaystyle \therefore AC^2=AD^2+(BC-BD)^2
\displaystyle =AD^2+BC^2+BD^2-2BC\times BD
\displaystyle =(AD^2+BD^2)+BC^2-2BC\times BD
\displaystyle \therefore AC^2=AB^2+BC^2-2BC\times BD\qquad\text{[Using (i)]}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Prove that in any triangle, the sum of the squares of any two sides is equal}
\displaystyle \text{to twice the square of half the third side together with twice the square of the}
\displaystyle \text{median which bisects the third side.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AD\text{ be the median of }\triangle ABC\text{ to the side }BC.
\displaystyle \therefore BD=DC=\frac{1}{2}BC.
\displaystyle \text{Produce }BD\text{ to }E,\text{ if necessary, and draw }AE\perp BC.
\displaystyle \text{In }\triangle ADB,\ \angle ADB\text{ is obtuse and }AE\perp BD.
\displaystyle \therefore AB^2=AD^2+BD^2+2BD\times DE\qquad\ldots\text{(i)}
\displaystyle \text{In }\triangle ADC,\ \angle ADC\text{ is acute and }AE\perp DC.
\displaystyle \therefore AC^2=AD^2+DC^2-2DC\times DE
\displaystyle \text{Since }DC=BD,
\displaystyle AC^2=AD^2+BD^2-2BD\times DE\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle AB^2+AC^2=2AD^2+2BD^2
\displaystyle =2AD^2+2\left(\frac{BC}{2}\right)^2
\displaystyle =2AD^2+\frac{1}{2}BC^2
\displaystyle \therefore AB^2+AC^2=2AD^2+2\left(\frac{BC}{2}\right)^2.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Prove that three times the sum of the squares of the sides of a triangle is}
\displaystyle \text{equal to four times the sum of the squares of its medians.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AD,\ BE\text{ and }CF\text{ be the medians of }\triangle ABC.
\displaystyle \therefore D,\ E\text{ and }F\text{ are the midpoints of }BC,\ CA\text{ and }AB,\text{ respectively.}
\displaystyle \text{Applying the median theorem to the median }AD,
\displaystyle AB^2+AC^2=2(AD^2+BD^2)
\displaystyle =2\left[AD^2+\left(\frac{BC}{2}\right)^2\right]
\displaystyle \therefore 2(AB^2+AC^2)=4AD^2+BC^2\qquad\ldots\text{(i)}
\displaystyle \text{Applying the median theorem to the median }BE,
\displaystyle AB^2+BC^2=2(BE^2+AE^2)
\displaystyle =2\left[BE^2+\left(\frac{AC}{2}\right)^2\right]
\displaystyle \therefore 2(AB^2+BC^2)=4BE^2+AC^2\qquad\ldots\text{(ii)}
\displaystyle \text{Applying the median theorem to the median }CF,
\displaystyle AC^2+BC^2=2(CF^2+AF^2)
\displaystyle =2\left[CF^2+\left(\frac{AB}{2}\right)^2\right]
\displaystyle \therefore 2(AC^2+BC^2)=4CF^2+AB^2\qquad\ldots\text{(iii)}
\displaystyle \text{Adding (i), (ii) and (iii), we get}
\displaystyle 4(AB^2+BC^2+CA^2)
\displaystyle =4(AD^2+BE^2+CF^2)+(AB^2+BC^2+CA^2)
\displaystyle \therefore 3(AB^2+BC^2+CA^2)=4(AD^2+BE^2+CF^2)
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }P\text{ and }Q\text{ are the midpoints of the sides }CA\text{ and }CB,\text{ respectively,}
\displaystyle \text{in }\triangle ABC.\text{ If }\angle C=90^\circ,\text{ prove that:}
\displaystyle \text{(i) }4AQ^2=4AC^2+BC^2\qquad\text{(ii) }4BP^2=4BC^2+AC^2
\displaystyle \text{(iii) }4(AQ^2+BP^2)=5AB^2. 2018-10-13_15-20-45
\displaystyle \text{Answer:}
\displaystyle \text{Since }P\text{ and }Q\text{ are the midpoints of }CA\text{ and }CB,\text{ respectively,}
\displaystyle CP=\frac{AC}{2}\quad\text{and}\quad CQ=\frac{BC}{2}.

\displaystyle \text{(i) In right-angled }\triangle AQC,\text{ by the Pythagoras theorem,}
\displaystyle AQ^2=AC^2+CQ^2
\displaystyle =AC^2+\left(\frac{BC}{2}\right)^2
\displaystyle \therefore 4AQ^2=4AC^2+BC^2.

\displaystyle \text{(ii) In right-angled }\triangle BPC,\text{ by the Pythagoras theorem,}
\displaystyle BP^2=BC^2+CP^2
\displaystyle =BC^2+\left(\frac{AC}{2}\right)^2
\displaystyle \therefore 4BP^2=4BC^2+AC^2.

\displaystyle \text{(iii) Adding the results obtained in (i) and (ii), we get}
\displaystyle 4AQ^2+4BP^2=(4AC^2+BC^2)+(4BC^2+AC^2)
\displaystyle 4(AQ^2+BP^2)=5(AC^2+BC^2)
\displaystyle \text{In right-angled }\triangle ABC,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AC^2+BC^2.
\displaystyle \therefore 4(AQ^2+BP^2)=5AB^2.
\displaystyle \therefore \text{All the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In the adjoining figure, }\triangle ABC\text{ is right-angled at }\angle B.
\displaystyle AD\text{ and }CE\text{ are medians drawn from the vertices }A\text{ and }C,\text{ respectively.}
\displaystyle \text{If }AC=5\text{ cm and }AD=\frac{3\sqrt5}{2}\text{ cm, find the length of }CE. 2018-10-13_15-20-54\displaystyle \text{Answer:}
\displaystyle \text{Since }AD\text{ and }CE\text{ are medians, }D\text{ and }E\text{ are the midpoints of }BC\text{ and }AB.
\displaystyle \therefore BD=\frac{BC}{2}\quad\text{and}\quad BE=\frac{AB}{2}.
\displaystyle \text{In right-angled }\triangle ABD,\text{ by the Pythagoras theorem,}
\displaystyle AD^2=AB^2+BD^2
\displaystyle =AB^2+\left(\frac{BC}{2}\right)^2
\displaystyle \therefore AD^2=AB^2+\frac14BC^2\qquad\ldots\text{(i)}
\displaystyle \text{In right-angled }\triangle BCE,\text{ by the Pythagoras theorem,}
\displaystyle CE^2=BC^2+BE^2
\displaystyle =BC^2+\left(\frac{AB}{2}\right)^2
\displaystyle \therefore CE^2=BC^2+\frac14AB^2\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle AD^2+CE^2=\frac54(AB^2+BC^2)
\displaystyle \text{In right-angled }\triangle ABC,\ AC^2=AB^2+BC^2.
\displaystyle \therefore AD^2+CE^2=\frac54AC^2
\displaystyle \left(\frac{3\sqrt5}{2}\right)^2+CE^2=\frac54(5)^2
\displaystyle \frac{45}{4}+CE^2=\frac{125}{4}
\displaystyle CE^2=\frac{125}{4}-\frac{45}{4}=\frac{80}{4}=20
\displaystyle \therefore CE=\sqrt{20}=2\sqrt5\text{ cm}.
\displaystyle \therefore \text{The length of }CE\text{ is }2\sqrt5\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In }\triangle ABC,\ \angle C=90^\circ.\text{ Let }BC=a,\ CA=b\text{ and }AB=c,
\displaystyle \text{and let }p\text{ be the length of the perpendicular drawn from }C\text{ to }AB.\text{ Prove that:}
\displaystyle \text{(i) }cp=ab
\displaystyle \text{(ii) }\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}.
\displaystyle \text{Answer:} 2018-10-13_15-20-21
\displaystyle \text{Let }CD\perp AB,\text{ where }CD=p.
\displaystyle \text{(i) Using }AB\text{ as the base of }\triangle ABC,
\displaystyle \text{Area of }\triangle ABC=\frac12\times AB\times CD=\frac12cp.
\displaystyle \text{Since }\angle C=90^\circ,\text{ using }BC\text{ as the base,}
\displaystyle \text{Area of }\triangle ABC=\frac12\times BC\times CA=\frac12ab.
\displaystyle \therefore \frac12cp=\frac12ab
\displaystyle \therefore cp=ab.

\displaystyle \text{(ii) In right-angled }\triangle ABC,\text{ by the Pythagoras theorem,}
\displaystyle c^2=a^2+b^2\qquad\ldots\text{(i)}
\displaystyle \text{From part (i), }cp=ab.
\displaystyle \text{Squaring both sides,}
\displaystyle c^2p^2=a^2b^2
\displaystyle \therefore c^2=\frac{a^2b^2}{p^2}
\displaystyle \text{Substituting this value of }c^2\text{ in (i), we get}
\displaystyle \frac{a^2b^2}{p^2}=a^2+b^2
\displaystyle \text{Dividing both sides by }a^2b^2,
\displaystyle \frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2}
\displaystyle \therefore \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}.
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In an equilateral triangle of side }a,\text{ prove that:}
\displaystyle \text{(i) Altitude}=\frac{a\sqrt3}{2}\qquad\text{(ii) Area}=\frac{\sqrt3}{4}a^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABC\text{ be an equilateral triangle of side }a\text{ and draw }AD\perp BC.
\displaystyle \text{Since the altitude of an equilateral triangle bisects the opposite side,}
\displaystyle BD=DC=\frac{BC}{2}=\frac{a}{2}.2018-10-13_15-20-33

\displaystyle \text{(i) In right-angled }\triangle ABD,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AD^2+BD^2
\displaystyle a^2=AD^2+\left(\frac{a}{2}\right)^2
\displaystyle AD^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}
\displaystyle \therefore AD=\frac{a\sqrt3}{2}.
\displaystyle \therefore \text{The altitude of the equilateral triangle is }\frac{a\sqrt3}{2}.

\displaystyle \text{(ii) Area of }\triangle ABC=\frac12\times BC\times AD
\displaystyle =\frac12\times a\times\frac{a\sqrt3}{2}
\displaystyle =\frac{\sqrt3}{4}a^2.
\displaystyle \therefore \text{The area of the equilateral triangle is }\frac{\sqrt3}{4}a^2.
\displaystyle \\

\displaystyle \textbf{Question 24: }ABC\text{ is a triangle in which }AB=AC\text{ and }D\text{ is any point on }BC.
\displaystyle \text{Prove that }AB^2-AD^2=BD\cdot CD.
\displaystyle \text{Answer:} 2018-10-13_15-19-51
\displaystyle \text{Draw }AE\perp BC.
\displaystyle \text{In right-angled triangles }AEB\text{ and }AEC,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle AE=AE\qquad\text{(Common)}
\displaystyle \angle AEB=\angle AEC=90^\circ
\displaystyle \therefore \triangle AEB\cong\triangle AEC\qquad\text{(RHS congruence criterion)}
\displaystyle \therefore BE=CE
\displaystyle \therefore BC=BE+CE=2BE\qquad\ldots\text{(i)}
\displaystyle \text{In right-angled }\triangle AEB,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AE^2+BE^2\qquad\ldots\text{(ii)}
\displaystyle \text{In right-angled }\triangle AED,\text{ by the Pythagoras theorem,}
\displaystyle AD^2=AE^2+DE^2\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting (iii) from (ii), we get}
\displaystyle AB^2-AD^2=BE^2-DE^2
\displaystyle \text{Since }DE^2=(BE-BD)^2\text{ for every position of }D\text{ on }BC,
\displaystyle AB^2-AD^2=BE^2-(BE-BD)^2
\displaystyle =\{BE-(BE-BD)\}\{BE+(BE-BD)\}
\displaystyle =BD(2BE-BD)
\displaystyle =BD(BC-BD)\qquad\text{[Using (i)]}
\displaystyle \text{Since }CD=BC-BD,
\displaystyle \therefore AB^2-AD^2=BD\cdot CD.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{From a point }O\text{ in the interior of }\triangle ABC,\text{ perpendiculars }OD,\ OE
\displaystyle \text{and }OF\text{ are drawn to the sides }BC,\ CA\text{ and }AB,\text{ respectively. Prove that:}
\displaystyle \text{(i) }AF^2+BD^2+CE^2=OA^2+OB^2+OC^2-OD^2-OE^2-OF^2
\displaystyle \text{(ii) }AF^2+BD^2+CE^2=AE^2+CD^2+BF^2.
\displaystyle \text{Answer:} 2018-10-13_15-20-03
\displaystyle \text{Given }OD\perp BC,\ OE\perp CA\text{ and }OF\perp AB.
\displaystyle \text{(i) In right-angled }\triangle AOF,\text{ by the Pythagoras theorem,}
\displaystyle OA^2=AF^2+OF^2
\displaystyle \text{In right-angled }\triangle BOD,
\displaystyle OB^2=BD^2+OD^2
\displaystyle \text{In right-angled }\triangle COE,
\displaystyle OC^2=CE^2+OE^2
\displaystyle \text{Adding the above three equations, we get}
\displaystyle OA^2+OB^2+OC^2=AF^2+BD^2+CE^2+OF^2+OD^2+OE^2
\displaystyle \therefore AF^2+BD^2+CE^2
\displaystyle =OA^2+OB^2+OC^2-OD^2-OE^2-OF^2.

\displaystyle \text{(ii) In right-angled triangles }ODB\text{ and }ODC,
\displaystyle OB^2=OD^2+BD^2
\displaystyle OC^2=OD^2+CD^2
\displaystyle \therefore OB^2-OC^2=BD^2-CD^2\qquad\ldots\text{(i)}
\displaystyle \text{In right-angled triangles }OEC\text{ and }OEA,
\displaystyle OC^2=OE^2+CE^2
\displaystyle OA^2=OE^2+AE^2
\displaystyle \therefore OC^2-OA^2=CE^2-AE^2\qquad\ldots\text{(ii)}
\displaystyle \text{In right-angled triangles }OFA\text{ and }OFB,
\displaystyle OA^2=OF^2+AF^2
\displaystyle OB^2=OF^2+BF^2
\displaystyle \therefore OA^2-OB^2=AF^2-BF^2\qquad\ldots\text{(iii)}
\displaystyle \text{Adding (i), (ii) and (iii), we get}
\displaystyle (OB^2-OC^2)+(OC^2-OA^2)+(OA^2-OB^2)
\displaystyle =(BD^2-CD^2)+(CE^2-AE^2)+(AF^2-BF^2)
\displaystyle 0=BD^2+CE^2+AF^2-CD^2-AE^2-BF^2
\displaystyle \therefore AF^2+BD^2+CE^2=AE^2+CD^2+BF^2.
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{A point }O\text{ inside a rectangle }ABCD\text{ is joined to each of the vertices}
\displaystyle A,\ B,\ C\text{ and }D.\text{ Prove that }OB^2+OD^2=OC^2+OA^2.
\displaystyle \text{Answer:}
\displaystyle \text{Through }O,\text{ draw }EF\parallel AB,\text{ meeting }AD\text{ at }E\text{ and }BC\text{ at }F.
\displaystyle \text{In right-angled }\triangle OEA,\text{ by the Pythagoras theorem,}
\displaystyle OA^2=OE^2+AE^2\qquad\ldots\text{(i)}2018-10-13_15-19-22
\displaystyle \text{In right-angled }\triangle OCF,
\displaystyle OC^2=OF^2+CF^2\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle OA^2+OC^2=OE^2+OF^2+AE^2+CF^2\qquad\ldots\text{(iii)}
\displaystyle \text{In right-angled }\triangle OFB,
\displaystyle OB^2=OF^2+BF^2\qquad\ldots\text{(iv)}
\displaystyle \text{In right-angled }\triangle OED,
\displaystyle OD^2=OE^2+DE^2\qquad\ldots\text{(v)}
\displaystyle \text{Adding (iv) and (v), we get}
\displaystyle OB^2+OD^2=OE^2+OF^2+BF^2+DE^2\qquad\ldots\text{(vi)}
\displaystyle \text{Since }ABCD\text{ is a rectangle and }EF\parallel AB,
\displaystyle AE=BF\quad\text{and}\quad DE=CF.
\displaystyle \therefore BF^2=AE^2\quad\text{and}\quad DE^2=CF^2.
\displaystyle \text{From (vi),}
\displaystyle OB^2+OD^2=OE^2+OF^2+AE^2+CF^2
\displaystyle \therefore OB^2+OD^2=OA^2+OC^2\qquad\text{[Using (iii)]}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

2018-10-13_15-19-36

\displaystyle \text{Question 27:  } \text{In }   \triangle ABC, AC > AB, D is mid point of \displaystyle  BC and \displaystyle  AE \perp BC . Prove:
i) \displaystyle  AC^2 = AD^2 + BC.DE + \frac{1}{2} BC^2
ii) \displaystyle  AB^2 = AD^2 - BC.DE + \frac{1}{4} BC^2
iii) \displaystyle  AB^2 + AC^2 = 2 AD^2 + \frac{1}{2} BC^2
Answer:
Given \displaystyle  \angle AED = 90^o
\displaystyle  \therefore \angle ADE < 90^o and \displaystyle  \angle ADC > 90^o
i) In \displaystyle  \triangle ADC, \angle ADC is an obtuse angle
\displaystyle  \therefore AC^2 = AD^2 + DC^2 + 2 DC.DE
\displaystyle  \Rightarrow AC^2 = AD^2 + ( \frac{1}{2} BC)^2 + 2 . \frac{1}{2} BC .DE
\displaystyle  \Rightarrow AC^2 = AD^2 + \frac{1}{4} BC^2 + BC.DE
\displaystyle  \Rightarrow AC^2 =AD^2 + BC.DE + \frac{1}{4} BC^2
ii) In \displaystyle  \triangle ABD, \angle ADE is an acute angle
\displaystyle  \therefore AB^2 = AD^2 + BD^2 - 2 BD .DE
\displaystyle  \Rightarrow AB^2 = AD^2 + (\frac{1}{2} BC)^2 + 2 . \frac{1}{2} BC.DE
\displaystyle  \Rightarrow AB^2 = AD^2 + \frac{1}{4} BC^2 - BC .DE
\displaystyle  \Rightarrow AB^2 = AD^2 - BC.DE + \frac{1}{4} BC^2
iii) From i) and ii) we get
\displaystyle  AB^2 + AC^2 = 2 AD^2 + \frac{1}{2} BC^2
\\

\displaystyle \textbf{Question 28: }\text{In }\triangle ABC,\ AC>AB,\ D\text{ is the midpoint of }BC\text{ and }AE\perp BC.
\displaystyle \text{Prove that:}
\displaystyle \text{(i) }AC^2=AD^2+BC\cdot DE+\frac14BC^2
\displaystyle \text{(ii) }AB^2=AD^2-BC\cdot DE+\frac14BC^2
\displaystyle \text{(iii) }AB^2+AC^2=2AD^2+\frac12BC^2.
\displaystyle \text{Answer:} 2018-10-13_15-19-51
\displaystyle \text{Since }D\text{ is the midpoint of }BC,
\displaystyle BD=DC=\frac12BC.

\displaystyle \text{(i) In right-angled }\triangle AEC,\text{ by the Pythagoras theorem,}
\displaystyle AC^2=AE^2+EC^2
\displaystyle \text{Since }EC=ED+DC,
\displaystyle AC^2=AE^2+(ED+DC)^2
\displaystyle =AE^2+ED^2+DC^2+2ED\cdot DC
\displaystyle \text{In right-angled }\triangle AED,\ AD^2=AE^2+ED^2.
\displaystyle \therefore AC^2=AD^2+DC^2+2ED\cdot DC
\displaystyle =AD^2+\left(\frac{BC}{2}\right)^2+2DE\left(\frac{BC}{2}\right)
\displaystyle \therefore AC^2=AD^2+BC\cdot DE+\frac14BC^2.

\displaystyle \text{(ii) In right-angled }\triangle AEB,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AE^2+BE^2
\displaystyle \text{Since }BE=BD-DE,
\displaystyle AB^2=AE^2+(BD-DE)^2
\displaystyle =AE^2+BD^2+DE^2-2BD\cdot DE
\displaystyle =(AE^2+DE^2)+BD^2-2BD\cdot DE
\displaystyle =AD^2+\left(\frac{BC}{2}\right)^2-2\left(\frac{BC}{2}\right)DE
\displaystyle \therefore AB^2=AD^2-BC\cdot DE+\frac14BC^2.

\displaystyle \text{(iii) Adding the results obtained in (i) and (ii), we get}
\displaystyle AB^2+AC^2
\displaystyle =AD^2-BC\cdot DE+\frac14BC^2+AD^2+BC\cdot DE+\frac14BC^2
\displaystyle \therefore AB^2+AC^2=2AD^2+\frac12BC^2.
\displaystyle \therefore \text{All the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{In the adjoining figure, }\triangle ABC\text{ is right-angled at }B.\text{ The points}
\displaystyle D\text{ and }E\text{ trisect }BC.\text{ Prove that }8AE^2=3AC^2+5AD^2. 2018-10-13_15-18-52\displaystyle \text{Answer:}
\displaystyle \text{Since }D\text{ and }E\text{ trisect }BC,
\displaystyle BD=DE=EC.
\displaystyle \text{Let }BD=DE=EC=x.
\displaystyle \therefore BE=2x\quad\text{and}\quad BC=3x.
\displaystyle \text{In right-angled }\triangle ABD,\text{ by the Pythagoras theorem,}
\displaystyle AD^2=AB^2+BD^2
\displaystyle \therefore AD^2=AB^2+x^2\qquad\ldots\text{(i)}
\displaystyle \text{In right-angled }\triangle ABE,\text{ by the Pythagoras theorem,}
\displaystyle AE^2=AB^2+BE^2
\displaystyle \therefore AE^2=AB^2+4x^2\qquad\ldots\text{(ii)}
\displaystyle \text{In right-angled }\triangle ABC,\text{ by the Pythagoras theorem,}
\displaystyle AC^2=AB^2+BC^2
\displaystyle \therefore AC^2=AB^2+9x^2\qquad\ldots\text{(iii)}
\displaystyle \text{Now,}
\displaystyle 8AE^2-3AC^2-5AD^2
\displaystyle =8(AB^2+4x^2)-3(AB^2+9x^2)-5(AB^2+x^2)
\displaystyle =8AB^2+32x^2-3AB^2-27x^2-5AB^2-5x^2
\displaystyle =0
\displaystyle \therefore 8AE^2=3AC^2+5AD^2.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\triangle ABC\text{ is a right-angled triangle with }\angle A=90^\circ.\text{ A circle is}
\displaystyle \text{inscribed in it. The lengths of the sides containing the right angle are }6\text{ cm and}
\displaystyle 8\text{ cm. Find the radius of the circle.}
\displaystyle \text{Answer:} 2018-10-13_15-18-24
\displaystyle \text{Let }AB=6\text{ cm},\ AC=8\text{ cm and }BC\text{ be the hypotenuse.}
\displaystyle \text{In right-angled }\triangle ABC,\text{ by the Pythagoras theorem,}
\displaystyle BC^2=AB^2+AC^2
\displaystyle =6^2+8^2=36+64=100
\displaystyle \therefore BC=10\text{ cm}.
\displaystyle \text{Let }O\text{ be the centre and }r\text{ be the radius of the inscribed circle.}
\displaystyle \text{The perpendicular distance of }O\text{ from each side of the triangle is }r.
\displaystyle \text{Area of }\triangle ABC
\displaystyle =\text{Area of }\triangle OAB+\text{Area of }\triangle OBC+\text{Area of }\triangle OCA
\displaystyle \frac12\times AB\times AC
\displaystyle =\frac12\times AB\times r+\frac12\times BC\times r+\frac12\times CA\times r
\displaystyle \frac12\times6\times8=\frac12r(6+10+8)
\displaystyle 24=12r
\displaystyle \therefore r=2\text{ cm}.
\displaystyle \therefore \text{The radius of the inscribed circle is }2\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{In the adjoining figure, }D\text{ is the midpoint of }BC\text{ and }AE\perp BC.
\displaystyle \text{If }BC=a,\ AC=b,\ AB=c,\ ED=x,\ AD=p\text{ and }AE=h,\text{ prove that:}
\displaystyle \text{(i) }b^2=p^2+ax+\frac{a^2}{4}
\displaystyle \text{(ii) }c^2=p^2-ax+\frac{a^2}{4}
\displaystyle \text{(iii) }b^2+c^2=2p^2+\frac{a^2}{2}. 2018-10-14_10-07-44.jpg\displaystyle \text{Answer:}
\displaystyle \text{Since }D\text{ is the midpoint of }BC,
\displaystyle BD=DC=\frac{BC}{2}=\frac a2.
\displaystyle \text{Also, }BE=BD-DE=\frac a2-x\text{ and }EC=ED+DC=x+\frac a2.
\displaystyle \text{In right-angled }\triangle AED,\text{ by the Pythagoras theorem,}
\displaystyle AD^2=AE^2+ED^2
\displaystyle p^2=h^2+x^2
\displaystyle \therefore h^2=p^2-x^2\qquad\ldots\text{(1)}

\displaystyle \text{(i) In right-angled }\triangle AEC,\text{ by the Pythagoras theorem,}
\displaystyle AC^2=AE^2+EC^2
\displaystyle b^2=h^2+\left(x+\frac a2\right)^2
\displaystyle =p^2-x^2+x^2+ax+\frac{a^2}{4}\qquad\text{[Using (1)]}
\displaystyle \therefore b^2=p^2+ax+\frac{a^2}{4}.

\displaystyle \text{(ii) In right-angled }\triangle AEB,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AE^2+BE^2
\displaystyle c^2=h^2+\left(\frac a2-x\right)^2
\displaystyle =p^2-x^2+\frac{a^2}{4}-ax+x^2\qquad\text{[Using (1)]}
\displaystyle \therefore c^2=p^2-ax+\frac{a^2}{4}.

\displaystyle \text{(iii) Adding the results obtained in (i) and (ii), we get}
\displaystyle b^2+c^2=p^2+ax+\frac{a^2}{4}+p^2-ax+\frac{a^2}{4}
\displaystyle \therefore b^2+c^2=2p^2+\frac{a^2}{2}.
\displaystyle \therefore \text{All the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{In the adjoining figure, }\angle B<90^\circ\text{ and }AD\perp BC.\text{ If }BC=a,
\displaystyle AC=b,\ AB=c,\ AD=h\text{ and }BD=x,\text{ prove that:}
\displaystyle \text{(i) }b^2=h^2+a^2+x^2-2ax
\displaystyle \text{(ii) }b^2=a^2+c^2-2ax. 2018-10-13_15-17-44\displaystyle \text{Answer:}
\displaystyle \text{Since }BC=a\text{ and }BD=x,
\displaystyle DC=BC-BD=a-x.
\displaystyle \text{In right-angled }\triangle ADB,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AD^2+BD^2
\displaystyle c^2=h^2+x^2
\displaystyle \therefore h^2=c^2-x^2\qquad\ldots\text{(1)}

\displaystyle \text{(i) In right-angled }\triangle ADC,\text{ by the Pythagoras theorem,}
\displaystyle AC^2=AD^2+DC^2
\displaystyle b^2=h^2+(a-x)^2
\displaystyle =h^2+a^2+x^2-2ax
\displaystyle \therefore b^2=h^2+a^2+x^2-2ax.

\displaystyle \text{(ii) Substituting }h^2=c^2-x^2\text{ from (1) in the result obtained in part (i),}
\displaystyle b^2=c^2-x^2+a^2+x^2-2ax
\displaystyle \therefore b^2=a^2+c^2-2ax.
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\triangle ABD\text{ is right-angled at }A\text{ and }AC\perp BD,\text{ where }C
\displaystyle \text{lies on }BD.\text{ Prove that:}
\displaystyle \text{(i) }AB^2=BC\cdot BD\qquad\text{(ii) }AC^2=BC\cdot DC
\displaystyle \text{(iii) }AD^2=BD\cdot CD\qquad\text{(iv) }\frac{AB^2}{AC^2}=\frac{BD}{DC}.
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,\text{ by the Pythagoras theorem,}
\displaystyle AB^2=AC^2+BC^2\qquad\ldots\text{(1)}
\displaystyle \text{In right-angled }\triangle ACD,\text{ by the Pythagoras theorem,}
\displaystyle AD^2=AC^2+DC^2\qquad\ldots\text{(2)}
\displaystyle \text{In right-angled }\triangle ABD,\text{ by the Pythagoras theorem,}
\displaystyle BD^2=AB^2+AD^2\qquad\ldots\text{(3)}

\displaystyle \text{(ii) Substituting (1) and (2) in (3), we get}
\displaystyle BD^2=AC^2+BC^2+AC^2+DC^2
\displaystyle BD^2=2AC^2+BC^2+DC^2
\displaystyle \text{Since }BD=BC+DC,
\displaystyle (BC+DC)^2=2AC^2+BC^2+DC^2
\displaystyle BC^2+DC^2+2BC\cdot DC=2AC^2+BC^2+DC^2
\displaystyle 2BC\cdot DC=2AC^2
\displaystyle \therefore AC^2=BC\cdot DC.

\displaystyle \text{(i) From (1),}
\displaystyle AB^2=AC^2+BC^2
\displaystyle =BC\cdot DC+BC^2
\displaystyle =BC(DC+BC)
\displaystyle \therefore AB^2=BC\cdot BD.

\displaystyle \text{(iii) From (2),}
\displaystyle AD^2=AC^2+DC^2
\displaystyle =BC\cdot DC+DC^2
\displaystyle =DC(BC+DC)
\displaystyle \therefore AD^2=DC\cdot BD.

\displaystyle \text{(iv) Dividing the result obtained in (i) by that obtained in (ii), we get}
\displaystyle \frac{AB^2}{AC^2}=\frac{BC\cdot BD}{BC\cdot DC}
\displaystyle \therefore \frac{AB^2}{AC^2}=\frac{BD}{DC}.
\displaystyle \therefore \text{All the required results are proved.}
\displaystyle \\

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