\displaystyle \textbf{Exercise - 15(A)}

\displaystyle \textbf{Question 1: }\text{In the given figure, }ABCD\text{ is a parallelogram in which}
\displaystyle \angle A=70^\circ.\text{ Calculate }\angle B,\angle C\text{ and }\angle D. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }ABCD\text{ is a parallelogram,}
\displaystyle \angle A=\angle C\text{ and }\angle B=\angle D.
\displaystyle \therefore \angle C=\angle A=70^\circ.
\displaystyle \text{Also, }\angle A+\angle B=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore 70^\circ+\angle B=180^\circ.
\displaystyle \Rightarrow \angle B=180^\circ-70^\circ=110^\circ.
\displaystyle \text{But }\angle D=\angle B.
\displaystyle \therefore \angle D=110^\circ.
\displaystyle {\therefore \angle B=110^\circ,\ \angle C=70^\circ\text{ and }\angle D=110^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In the given figure, }ABCD\text{ is a parallelogram. Side }DC\text{ is}
\displaystyle \text{produced to }E\text{ and }\angle BCE=105^\circ.\text{ Calculate }\angle A,\angle B,\angle C\text{ and }\angle D. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }ABCD\text{ is a parallelogram and side }DC\text{ is produced to }E,
\displaystyle \angle BCE+\angle BCD=180^\circ.
\displaystyle \text{(Linear pair)}
\displaystyle \therefore 105^\circ+\angle BCD=180^\circ.
\displaystyle \Rightarrow \angle BCD=180^\circ-105^\circ=75^\circ.
\displaystyle \therefore \angle C=75^\circ.
\displaystyle \text{But }\angle A=\angle C.
\displaystyle \text{(Opposite angles of a parallelogram)}
\displaystyle \therefore \angle A=75^\circ.
\displaystyle \text{Also, }AB\parallel CD.
\displaystyle \therefore \angle BCE=\angle CBA.
\displaystyle \text{(Alternate angles)}
\displaystyle \therefore \angle CBA=105^\circ\text{ or }\angle B=105^\circ.
\displaystyle \text{But }\angle D=\angle B.
\displaystyle \text{(Opposite angles of a parallelogram)}
\displaystyle \therefore \angle D=105^\circ.
\displaystyle {\therefore \angle A=75^\circ,\ \angle B=105^\circ,\ \angle C=75^\circ\text{ and }\angle D=105^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If an angle of a parallelogram is two-third of its adjacent angle,}
\displaystyle \text{find the angles of the parallelogram.} \displaystyle \textbf{Answer:}
\displaystyle \text{Let }\angle B=x.
\displaystyle \therefore \angle A=\frac{2}{3}x.
\displaystyle \text{But }\angle A+\angle B=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore \frac{2}{3}x+x=180^\circ.
\displaystyle \Rightarrow \frac{5}{3}x=180^\circ.
\displaystyle \Rightarrow x=180^\circ\times\frac{3}{5}=108^\circ.
\displaystyle \therefore \angle B=108^\circ\text{ and }\angle A=\frac{2}{3}\times108^\circ=72^\circ.
\displaystyle \text{Also, }\angle C=\angle A=72^\circ\text{ and }\angle D=\angle B=108^\circ.
\displaystyle {\therefore \angle A=72^\circ,\ \angle B=108^\circ,\ \angle C=72^\circ\text{ and }\angle D=108^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram in which}
\displaystyle \angle BAD=70^\circ\text{ and }\angle CBD=50^\circ.\text{ Calculate:}
\displaystyle \text{(i) }\angle ADB \hspace{1cm}\text{(ii) }\angle CDB \displaystyle \textbf{Answer:}
\displaystyle \text{Since }ABCD\text{ is a parallelogram, }AD\parallel BC.
\displaystyle \text{Also, }BD\text{ is a transversal.}
\displaystyle \therefore \angle ADB=\angle CBD=50^\circ.
\displaystyle \text{(Alternate angles)}
\displaystyle \text{Also, }\angle BAD+\angle ADC=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore \angle BAD+\angle ADB+\angle CDB=180^\circ.
\displaystyle \Rightarrow 70^\circ+50^\circ+\angle CDB=180^\circ.
\displaystyle \Rightarrow \angle CDB=180^\circ-120^\circ=60^\circ.
\displaystyle {\therefore \text{(i) }\angle ADB=50^\circ\text{ and (ii) }\angle CDB=60^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the given figure, }ABCD\text{ is a rhombus in which }\angle A=72^\circ.
\displaystyle \text{If }\angle CBD=x^\circ,\text{ find the value of }x. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }ABCD\text{ is a rhombus, diagonal }BD\text{ bisects }\angle B\text{ and }\angle D.
\displaystyle \therefore \angle ABD=\angle CBD=x^\circ.
\displaystyle \therefore \angle ABC=x+x=2x.
\displaystyle \text{Also, }\angle A+\angle B=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore 72^\circ+2x=180^\circ.
\displaystyle \Rightarrow 2x=180^\circ-72^\circ=108^\circ.
\displaystyle \therefore x=\frac{108^\circ}{2}=54^\circ.
\displaystyle {\therefore \angle CBD=54^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the adjoining figure, equilateral }\triangle EDC\text{ surmounts square }ABCD.
\displaystyle \text{If }\angle DEB=x^\circ,\text{ find the value of }x. \displaystyle \textbf{Answer:}
\displaystyle \text{In the figure, }ABCD\text{ is a square and }\triangle CDE\text{ is equilateral.}
\displaystyle \text{Also, }BE\text{ is joined and }\angle DEB=x^\circ.
\displaystyle \text{In }\triangle BCE,\ BC=CE.
\displaystyle \therefore \angle CBE=\angle CEB.
\displaystyle \text{Also, }\angle BCE=\angle BCD+\angle DCE.
\displaystyle =90^\circ+60^\circ=150^\circ.
\displaystyle \text{But }\angle BCE+\angle CBE+\angle CEB=180^\circ.
\displaystyle \text{(Angles of a triangle)}
\displaystyle \therefore 150^\circ+\angle CEB+\angle CEB=180^\circ.
\displaystyle \Rightarrow 150^\circ+2\angle CEB=180^\circ.
\displaystyle \Rightarrow 2\angle CEB=30^\circ.
\displaystyle \therefore \angle CEB=\frac{30^\circ}{2}=15^\circ.
\displaystyle \text{But }\angle CED=60^\circ.
\displaystyle \text{(Angle of an equilateral triangle)}
\displaystyle \therefore x^\circ+\angle CEB=60^\circ.
\displaystyle \Rightarrow x^\circ+15^\circ=60^\circ.
\displaystyle \therefore x=60^\circ-15^\circ=45^\circ.
\displaystyle {\therefore \text{The value of }x\text{ is }45^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In the adjoining figure, }ABCD\text{ is a rhombus whose diagonals}
\displaystyle \text{intersect at }O.\text{ If }\angle OAB:\angle OBA=2:3,\text{ find the angles of }\triangle OAB. \displaystyle \textbf{Answer:}
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \therefore \angle AOB=90^\circ.
\displaystyle \text{Given }\angle OAB:\angle OBA=2:3.
\displaystyle \text{Let }\angle OAB=2x\text{ and }\angle OBA=3x.
\displaystyle \text{In }\triangle OAB,
\displaystyle \angle OAB+\angle OBA+\angle AOB=180^\circ.
\displaystyle \therefore 2x+3x+90^\circ=180^\circ.
\displaystyle \Rightarrow 5x=90^\circ.
\displaystyle \therefore x=\frac{90^\circ}{5}=18^\circ.
\displaystyle \therefore \angle OAB=2x=2\times18^\circ=36^\circ.
\displaystyle \angle OBA=3x=3\times18^\circ=54^\circ.
\displaystyle {\therefore \angle OAB=36^\circ,\ \angle OBA=54^\circ\text{ and }\angle AOB=90^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the given figure, }ABCD\text{ is a rectangle whose diagonals intersect at }O.
\displaystyle \text{Diagonal }AC\text{ is produced to }E\text{ and }\angle ECD=140^\circ.
\displaystyle \text{Find the angles of }\triangle OAB. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }ABCD\text{ is a rectangle, its diagonals }AC\text{ and }BD\text{ bisect each other at }O.
\displaystyle \text{Diagonal }AC\text{ is produced to }E\text{ such that }\angle ECD=140^\circ.
\displaystyle \angle ECD+\angle DCO=180^\circ.
\displaystyle \text{(Linear pair)}
\displaystyle \therefore 140^\circ+\angle DCO=180^\circ.
\displaystyle \Rightarrow \angle DCO=180^\circ-140^\circ=40^\circ.
\displaystyle \text{Also, }OC=OD.
\displaystyle \text{(Halves of equal diagonals)}
\displaystyle \therefore \angle CDO=\angle DCO=40^\circ.
\displaystyle \text{Now, }AB\parallel CD.
\displaystyle \text{(Opposite sides of a rectangle)}
\displaystyle \therefore \angle OAB=\angle DCO=40^\circ.
\displaystyle \text{(Alternate angles)}
\displaystyle \text{Similarly, }\angle OBA=40^\circ.
\displaystyle \text{In }\triangle AOB,
\displaystyle \angle OBA+\angle OAB+\angle AOB=180^\circ.
\displaystyle \text{(Angles of a triangle)}
\displaystyle \therefore 40^\circ+40^\circ+\angle AOB=180^\circ.
\displaystyle \Rightarrow 80^\circ+\angle AOB=180^\circ.
\displaystyle \therefore \angle AOB=180^\circ-80^\circ=100^\circ.
\displaystyle {\therefore \angle OAB=40^\circ,\ \angle OBA=40^\circ\text{ and }\angle AOB=100^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the given figure, }ABCD\text{ is a kite whose diagonals intersect at }O.
\displaystyle \text{If }\angle DAB=54^\circ\text{ and }\angle BCD=76^\circ,\text{ calculate:}
\displaystyle \text{(i) }\angle ODA \hspace{1cm}\text{(ii) }\angle OBC. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }ABCD\text{ is a kite, }AB=AD\text{ and }BC=DC.
\displaystyle \text{Its diagonals }AC\text{ and }BD\text{ intersect at }O.
\displaystyle \text{Given }\angle DAB=54^\circ\text{ and }\angle BCD=76^\circ.
\displaystyle \text{In }\triangle BCD,\ BC=DC.
\displaystyle \therefore \angle CDB=\angle CBD.
\displaystyle \text{But }\angle BCD+\angle CDB+\angle CBD=180^\circ.
\displaystyle \therefore 76^\circ+\angle CBD+\angle CDB=180^\circ.
\displaystyle \Rightarrow 76^\circ+2\angle CBD=180^\circ.
\displaystyle \Rightarrow 2\angle CBD=180^\circ-76^\circ=104^\circ.
\displaystyle \therefore \angle CBD=\frac{104^\circ}{2}=52^\circ.
\displaystyle \therefore \angle OBC=52^\circ.
\displaystyle \text{Similarly, in }\triangle ABD,\ AB=AD.
\displaystyle \therefore \angle ABD=\angle ADB.
\displaystyle \text{But }\angle DAB+\angle ABD+\angle ADB=180^\circ.
\displaystyle \therefore 54^\circ+\angle ADB+\angle ADB=180^\circ.
\displaystyle \Rightarrow 54^\circ+2\angle ADB=180^\circ.
\displaystyle \Rightarrow 2\angle ADB=180^\circ-54^\circ=126^\circ.
\displaystyle \therefore \angle ADB=\frac{126^\circ}{2}=63^\circ.
\displaystyle \therefore \angle ODA=63^\circ.
\displaystyle {\therefore \text{(i) }\angle ODA=63^\circ\text{ and (ii) }\angle OBC=52^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the given figure, }ABCD\text{ is an isosceles trapezium in which}
\displaystyle \angle CDA=2x^\circ\text{ and }\angle BAD=3x^\circ.
\displaystyle \text{Find all the angles of the trapezium.} \displaystyle \textbf{Answer:}
\displaystyle \text{In isosceles trapezium }ABCD,\ AD=BC\text{ and }AB\parallel CD.
\displaystyle \angle BAD+\angle CDA=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore 3x+2x=180^\circ.
\displaystyle \Rightarrow 5x=180^\circ.
\displaystyle \therefore x=\frac{180^\circ}{5}=36^\circ.
\displaystyle \therefore \angle A=3x=3\times36^\circ=108^\circ.
\displaystyle \angle D=2x=2\times36^\circ=72^\circ.
\displaystyle \text{Since }ABCD\text{ is an isosceles trapezium,}
\displaystyle \angle A=\angle B\text{ and }\angle C=\angle D.
\displaystyle \therefore \angle B=108^\circ\text{ and }\angle C=72^\circ.
\displaystyle {\therefore \angle A=108^\circ,\ \angle B=108^\circ,\ \angle C=72^\circ\text{ and }\angle D=72^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the given figure, }ABCD\text{ is a trapezium in which}
\displaystyle \angle A=(x+25)^\circ,\ \angle B=y^\circ,\ \angle C=95^\circ
\displaystyle \text{and }\angle D=(2x+5)^\circ.\text{ Find the values of }x\text{ and }y. \displaystyle \textbf{Answer:}
\displaystyle \text{In trapezium }ABCD,\ AB\parallel CD.
\displaystyle \angle A+\angle D=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore (x+25)^\circ+(2x+5)^\circ=180^\circ.
\displaystyle \Rightarrow x+25+2x+5=180.
\displaystyle \Rightarrow 3x+30=180.
\displaystyle \Rightarrow 3x=180-30=150.
\displaystyle \therefore x=\frac{150}{3}=50.
\displaystyle \text{Similarly, }\angle B+\angle C=180^\circ.
\displaystyle \therefore y^\circ+95^\circ=180^\circ.
\displaystyle \Rightarrow y=180-95=85.
\displaystyle {\therefore x=50\text{ and }y=85.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the given figure, }ABCD\text{ is a rhombus and }\triangle EDC\text{ is}
\displaystyle \text{equilateral. If }\angle BAD=78^\circ,\text{ calculate: (i) }\angle CBE\text{ (ii) }\angle DBE. \displaystyle \textbf{Answer:}
\displaystyle \textbf{(i)}\ ABCD\text{ is a rhombus and }\triangle EDC\text{ is an equilateral triangle.}
\displaystyle \angle BAD=78^\circ.
\displaystyle \therefore \angle BCD=\angle A=78^\circ.
\displaystyle \text{(Opposite angles of a rhombus)}
\displaystyle \therefore \angle BCE=\angle BCD+\angle DCE.
\displaystyle =78^\circ+60^\circ=138^\circ.
\displaystyle \text{Also, }BC=CE.
\displaystyle \therefore \angle CBE=\angle CEB.
\displaystyle \text{But }\angle CBE+\angle CEB+\angle BCE=180^\circ.
\displaystyle \text{(Sum of the angles of a triangle)}
\displaystyle \therefore \angle CBE+\angle CBE+138^\circ=180^\circ.
\displaystyle \Rightarrow 2\angle CBE=180^\circ-138^\circ=42^\circ.
\displaystyle \therefore \angle CBE=\frac{42^\circ}{2}=21^\circ.
\displaystyle \textbf{(ii)}\text{ In }\triangle ABD,\ AB=AD.
\displaystyle \text{(Sides of a rhombus)}
\displaystyle \therefore \angle ABD=\angle ADB.
\displaystyle \text{But }\angle ABD+\angle ADB+\angle BAD=180^\circ.
\displaystyle \therefore \angle ABD+\angle ABD+78^\circ=180^\circ.
\displaystyle \Rightarrow 2\angle ABD=180^\circ-78^\circ=102^\circ.
\displaystyle \therefore \angle ABD=\frac{102^\circ}{2}=51^\circ.
\displaystyle \text{Also, }\angle BAD+\angle ABC=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore 78^\circ+\angle ABC=180^\circ.
\displaystyle \Rightarrow \angle ABC=180^\circ-78^\circ=102^\circ.
\displaystyle \therefore \angle ABD+\angle DBE+\angle CBE=102^\circ.
\displaystyle \Rightarrow 51^\circ+\angle DBE+21^\circ=102^\circ.
\displaystyle \Rightarrow 72^\circ+\angle DBE=102^\circ.
\displaystyle \therefore \angle DBE=102^\circ-72^\circ=30^\circ.
\displaystyle {\therefore \text{(i) }\angle CBE=21^\circ\text{ and (ii) }\angle DBE=30^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\triangle DEC\text{ is an equilateral triangle in a square }ABCD.
\displaystyle \text{If }BD\text{ and }CE\text{ intersect at }O\text{ and }\angle COD=x^\circ,\text{ find the value of }x. \displaystyle \textbf{Answer:}
\displaystyle ABCD\text{ is a square and }\triangle ECD\text{ is an equilateral triangle.}
\displaystyle \text{Diagonal }BD\text{ and }CE\text{ intersect each other at }O,\ \angle COD=x^\circ.
\displaystyle \text{Since }BD\text{ is a diagonal of square }ABCD,
\displaystyle \angle BDC=\frac{90^\circ}{2}=45^\circ.
\displaystyle \therefore \angle ODC=45^\circ.
\displaystyle \angle ECD=60^\circ.
\displaystyle \text{(Angle of an equilateral triangle)}
\displaystyle \therefore \angle OCD=60^\circ.
\displaystyle \text{In }\triangle OCD,
\displaystyle \angle OCD+\angle ODC+\angle COD=180^\circ.
\displaystyle \text{(Angles of a triangle)}
\displaystyle \therefore 60^\circ+45^\circ+x^\circ=180^\circ.
\displaystyle \Rightarrow 105^\circ+x^\circ=180^\circ.
\displaystyle \therefore x^\circ=180^\circ-105^\circ=75^\circ.
\displaystyle {\therefore \text{The value of }x\text{ is }75^\circ.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If one angle of a parallelogram is }90^\circ,\text{ show that each of its}
\displaystyle \text{angles measures }90^\circ. \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram and }\angle A=90^\circ.
\displaystyle \textbf{To prove: }\text{Each angle of parallelogram }ABCD\text{ is }90^\circ.
\displaystyle \textbf{Proof:}
\displaystyle \angle A=\angle C.
\displaystyle \text{(Opposite angles of a parallelogram)}
\displaystyle \therefore \angle C=90^\circ.
\displaystyle \text{But }\angle A+\angle D=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore \angle D=180^\circ-90^\circ=90^\circ.
\displaystyle \text{Also, }\angle B=\angle D.
\displaystyle \text{(Opposite angles of a parallelogram)}
\displaystyle \therefore \angle B=90^\circ.
\displaystyle {\therefore \angle A=\angle B=\angle C=\angle D=90^\circ.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the adjoining figure, }ABCD\text{ and }PQBA\text{ are two parallelograms.} \displaystyle \text{Prove that:}
\displaystyle \text{(i) }DPQC\text{ is a parallelogram.}
\displaystyle \text{(ii) }DP=CQ.
\displaystyle \text{(iii) }\triangle DAP\cong\triangle CBQ.
\displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ and }PQBA\text{ are two parallelograms. }DP\text{ and }QC\text{ are joined.}
\displaystyle \textbf{To prove: }\text{(i) }DPQC\text{ is a parallelogram, (ii) }DP=CQ
\displaystyle \text{and (iii) }\triangle DAP\cong\triangle CBQ.
\displaystyle \textbf{Proof:}
\displaystyle \textbf{(i)}\ DC\parallel AB\text{ and }AB\parallel PQ.
\displaystyle \text{(Given)}
\displaystyle \therefore DC\parallel PQ.
\displaystyle \text{Also, }DC=AB\text{ and }AB=PQ.
\displaystyle \text{(Opposite sides of parallelograms)}
\displaystyle \therefore DC=PQ.
\displaystyle \text{Thus, }DC=PQ\text{ and }DC\parallel PQ.
\displaystyle \therefore DPQC\text{ is a parallelogram.}
\displaystyle \textbf{(ii)}\ DP=CQ.
\displaystyle \text{(Opposite sides of parallelogram }DPQC\text{)}
\displaystyle \textbf{(iii)}\text{ In }\triangle DAP\text{ and }\triangle CBQ,
\displaystyle DA=CB.
\displaystyle \text{(Opposite sides of parallelogram }ABCD\text{)}
\displaystyle AP=BQ.
\displaystyle \text{(Opposite sides of parallelogram }PQBA\text{)}
\displaystyle DP=CQ.
\displaystyle \text{(Proved)}
\displaystyle \therefore \triangle DAP\cong\triangle CBQ.
\displaystyle \text{(S.S.S. axiom of congruency)}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram. }BM\perp AC
\displaystyle \text{and }DN\perp AC.\text{ Prove that: (i) }\triangle BMC\cong\triangle DNA
\displaystyle \text{(ii) }BM=DN. \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram, }BM\perp AC\text{ and }DN\perp AC.
\displaystyle \textbf{To prove: }\text{(i) }\triangle BMC\cong\triangle DNA\text{ and (ii) }BM=DN.
\displaystyle \textbf{Proof:}\text{ In }\triangle BMC\text{ and }\triangle DNA,
\displaystyle BC=AD.
\displaystyle \text{(Opposite sides of a parallelogram)}
\displaystyle \angle BMC=\angle DNA=90^\circ.
\displaystyle \angle BCM=\angle DAN.
\displaystyle \text{(Alternate angles, since }BC\parallel AD\text{)}
\displaystyle \therefore \triangle BMC\cong\triangle DNA.
\displaystyle \text{(A.A.S. axiom of congruency)}
\displaystyle \therefore BM=DN.
\displaystyle \text{(C.P.C.T.)}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram and }X\text{ is the}
\displaystyle \text{mid-point of }BC.\text{ The line }AX\text{ produced meets }DC\text{ produced at }Q.
\displaystyle \text{The parallelogram }AQPB\text{ is completed. Prove that:}
\displaystyle \text{(i) }\triangle ABX\cong\triangle QCX.
\displaystyle \text{(ii) }DC=CQ=QP. \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram and }X\text{ is the mid-point of }BC.
\displaystyle AX\text{ is produced to meet }DC\text{ produced at }Q.
\displaystyle BP\parallel AQ\text{ is drawn so that }AQPB\text{ is a parallelogram.}
\displaystyle \textbf{To prove: }\text{(i) }\triangle ABX\cong\triangle QCX\text{ and (ii) }DC=CQ=QP.
\displaystyle \textbf{Proof:}
\displaystyle \textbf{(i)}\text{ In }\triangle ABX\text{ and }\triangle QCX,
\displaystyle XB=XC.
\displaystyle \text{(Since }X\text{ is the mid-point of }BC\text{)}
\displaystyle \angle AXB=\angle CXQ.
\displaystyle \text{(Vertically opposite angles)}
\displaystyle \angle BAX=\angle XQC.
\displaystyle \text{(Alternate angles, since }AB\parallel DC\text{)}
\displaystyle \therefore \triangle ABX\cong\triangle QCX.
\displaystyle \text{(A.A.S. axiom of congruency)}
\displaystyle \textbf{(ii)}\text{ In parallelogram }ABCD,
\displaystyle AB=DC. \qquad \cdots(i)
\displaystyle \text{(Opposite sides of a parallelogram)}
\displaystyle \text{Similarly, in parallelogram }AQPB,
\displaystyle AB=QP. \qquad \cdots(ii)
\displaystyle \text{From }(i)\text{ and }(ii),
\displaystyle DC=QP. \qquad \cdots(iii)
\displaystyle \text{In }\triangle BCP,\ X\text{ is the mid-point of }BC\text{ and }XQ\parallel BP.
\displaystyle \therefore Q\text{ is the mid-point of }CP.
\displaystyle \therefore CQ=QP. \qquad \cdots(iv)
\displaystyle \text{From }(iii)\text{ and }(iv),
\displaystyle DC=QP=CQ.
\displaystyle \therefore DC=CQ=QP.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram.}
\displaystyle \text{Line segments }AX\text{ and }CY\text{ bisect }\angle A\text{ and }\angle C\text{ respectively.}
\displaystyle \text{Prove that:}
\displaystyle \text{(i) }\triangle ADX\cong\triangle CBY
\displaystyle \text{(ii) }AX=CY
\displaystyle \text{(iii) }AX\parallel CY
\displaystyle \text{(iv) }AYCX\text{ is a parallelogram.} \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram.}
\displaystyle \text{Line segments }AX\text{ and }CY\text{ bisect }\angle A\text{ and }\angle C\text{ respectively.}
\displaystyle \textbf{To prove: }\text{(i) }\triangle ADX\cong\triangle CBY,\text{ (ii) }AX=CY,
\displaystyle \text{(iii) }AX\parallel CY\text{ and (iv) }AYCX\text{ is a parallelogram.}
\displaystyle \textbf{Proof:}
\displaystyle \textbf{(i)}\text{ In }\triangle ADX\text{ and }\triangle CBY,
\displaystyle AD=BC.
\displaystyle \text{(Opposite sides of a parallelogram)}
\displaystyle \angle D=\angle B.
\displaystyle \text{(Opposite angles of a parallelogram)}
\displaystyle \angle DAX=\angle BCY.
\displaystyle \text{(Halves of equal angles }\angle A\text{ and }\angle C\text{)}
\displaystyle \therefore \triangle ADX\cong\triangle CBY.
\displaystyle \text{(A.A.S. axiom of congruency)}
\displaystyle \textbf{(ii)}\ AX=CY.
\displaystyle \text{(C.P.C.T.)}
\displaystyle \textbf{(iii)}\ \angle1=\angle2.
\displaystyle \text{(Halves of equal angles)}
\displaystyle \text{But }\angle2=\angle3.
\displaystyle \text{(Alternate angles, since }AB\parallel DC\text{)}
\displaystyle \therefore \angle1=\angle3.
\displaystyle \text{These are corresponding angles.}
\displaystyle \therefore AX\parallel CY.
\displaystyle \textbf{(iv)}\ AX=CY\text{ and }AX\parallel CY.
\displaystyle \therefore AYCX\text{ is a parallelogram.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In the given figure, }ABCD\text{ is a parallelogram and }X,Y\text{ are points}
\displaystyle \text{on diagonal }BD\text{ such that }DX=BY.\text{ Prove that }CXAY\text{ is a parallelogram.} \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram. }X\text{ and }Y\text{ are points on diagonal }BD
\displaystyle \text{such that }DX=BY.
\displaystyle \textbf{To prove: }CXAY\text{ is a parallelogram.}
\displaystyle \textbf{Construction: }AC\text{ is joined, meeting }BD\text{ at }O.
\displaystyle \textbf{Proof: }AC\text{ and }BD\text{ are the diagonals of parallelogram }ABCD.
\displaystyle \therefore AC\text{ and }BD\text{ bisect each other at }O.
\displaystyle \therefore AO=OC\text{ and }BO=OD.
\displaystyle \text{But }DX=BY.
\displaystyle \text{(Given)}
\displaystyle \therefore DO-DX=OB-BY.
\displaystyle \Rightarrow OX=OY.
\displaystyle \text{In quadrilateral }CXAY,\text{ diagonals }AC\text{ and }XY\text{ bisect each other at }O.
\displaystyle \therefore CXAY\text{ is a parallelogram.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Show that the bisectors of the angles of a parallelogram enclose}
\displaystyle \text{a rectangle.} \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram.}
\displaystyle \text{The bisectors of }\angle A\text{ and }\angle B\text{ meet at }S,\text{ and the bisectors of}
\displaystyle \angle C\text{ and }\angle D\text{ meet at }Q.
\displaystyle \textbf{To prove: }PQRS\text{ is a rectangle.}
\displaystyle \textbf{Proof: }\angle A+\angle B=180^\circ.
\displaystyle \therefore \frac{1}{2}\angle A+\frac{1}{2}\angle B=90^\circ.
\displaystyle \Rightarrow \angle SAB+\angle SBA=90^\circ.
\displaystyle \text{In }\triangle ASB,
\displaystyle \angle ASB=90^\circ.
\displaystyle \text{Similarly, }\angle CQD=90^\circ.
\displaystyle \text{Again, }\angle A+\angle D=180^\circ.
\displaystyle \text{(Co-interior angles)}
\displaystyle \therefore \frac{1}{2}\angle A+\frac{1}{2}\angle D=90^\circ.
\displaystyle \Rightarrow \angle PAD+\angle PDA=90^\circ.
\displaystyle \therefore \angle APD=90^\circ.
\displaystyle \text{But }\angle SPQ=\angle APD.
\displaystyle \text{(Vertically opposite angles)}
\displaystyle \therefore \angle SPQ=90^\circ.
\displaystyle \text{Similarly, }\angle SRQ=90^\circ.
\displaystyle \text{Thus, each angle of quadrilateral }PQRS\text{ is }90^\circ.
\displaystyle \therefore PQRS\text{ is a rectangle.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If a diagonal of a parallelogram bisects one of the angles of the}
\displaystyle \text{parallelogram, prove that it also bisects the second angle and that the two}
\displaystyle \text{diagonals are perpendicular to each other.} \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In parallelogram }ABCD,\text{ diagonal }AC\text{ bisects }\angle A.
\displaystyle BD\text{ is joined, meeting }AC\text{ at }O.
\displaystyle \textbf{To prove: }\text{(i) }AC\text{ bisects }\angle C.
\displaystyle \text{(ii) Diagonals }AC\text{ and }BD\text{ are perpendicular to each other.}
\displaystyle \textbf{Proof: }\text{Since }AB\parallel DC,
\displaystyle \angle1=\angle4\text{ and }\angle2=\angle3.
\displaystyle \text{(Alternate angles)}
\displaystyle \text{But }\angle1=\angle2.
\displaystyle \text{(Given)}
\displaystyle \therefore \angle3=\angle4.
\displaystyle \therefore AC\text{ bisects }\angle C.
\displaystyle \text{Similarly, diagonal }BD\text{ bisects }\angle B\text{ and }\angle D.
\displaystyle \therefore ABCD\text{ is a rhombus.}
\displaystyle \text{But the diagonals of a rhombus bisect each other at right angles.}
\displaystyle \therefore AC\perp BD.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In the given figure, }ABCD\text{ is a parallelogram and }E\text{ is the}
\displaystyle \text{mid-point of }BC.\text{ If }DE\text{ and }AB\text{ produced meet at }F,\text{ prove that }AF=2AB. \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram and }E\text{ is the mid-point of }BC.
\displaystyle DE\text{ and }AB\text{ produced meet at }F.
\displaystyle \textbf{To prove: }AF=2AB.
\displaystyle \textbf{Proof: }\text{In }\triangle DEC\text{ and }\triangle FEB,
\displaystyle CE=EB.
\displaystyle \text{(Since }E\text{ is the mid-point of }BC\text{)}
\displaystyle \angle DEC=\angle BEF.
\displaystyle \text{(Vertically opposite angles)}
\displaystyle \angle DCE=\angle EBF.
\displaystyle \text{(Alternate angles, since }DC\parallel AB\text{)}
\displaystyle \therefore \triangle DEC\cong\triangle FEB.
\displaystyle \text{(A.A.S. axiom of congruency)}
\displaystyle \therefore CD=BF.
\displaystyle \text{(C.P.C.T.)}
\displaystyle \text{But }AB=CD.
\displaystyle \text{(Opposite sides of a parallelogram)}
\displaystyle \therefore AB=BF.
\displaystyle \text{Now, }AF=AB+BF.
\displaystyle =AB+AB=2AB.
\displaystyle \therefore AF=2AB.
\displaystyle \text{Hence proved.}
\displaystyle \\

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