\displaystyle \textbf{Question 2: }\text{State, giving reasons, whether the following}
\displaystyle \text{pairs of triangles are congruent or not.}
\displaystyle \text{(i) }\Delta ABC\text{ in which }\angle A=50^\circ,\ \angle B=60^\circ,\ BC=4.5\text{ cm, and }\Delta DEF
\displaystyle \text{in which }\angle E=60^\circ,\ EF=4.5\text{ cm and } \angle F=70^\circ.
\displaystyle \text{(ii) }\Delta DEF\text{ in which }\angle E=48^\circ,\ DE=6\text{ cm},\ EF=8\text{ cm, and }\Delta MNR
\displaystyle \text{in which }\angle R=48^\circ,\ MN=6\text{ cm and } MR=8\text{ cm}.
\displaystyle \text{(iii) }\Delta KLM\text{ in which }KM=4\text{ cm},\ \angle K=75^\circ,\ \angle M=40^\circ,\text{ and }\Delta PQR
\displaystyle \text{in which }PR=4\text{ cm},\ \angle Q=65^\circ \text{ and }\angle R=40^\circ.
\displaystyle \text{(iv) }\Delta ABC\text{ in which }AB=3\text{ cm},\ \angle A=90^\circ,\ BC=5\text{ cm, and }\Delta KLM
\displaystyle \text{in which }KM=3\text{ cm},\ \angle K=90^\circ \text{ and }LM=5\text{ cm}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) In }\Delta ABC,\quad \angle C=180^\circ-50^\circ-60^\circ=70^\circ.
\displaystyle \text{In }\Delta DEF,\quad \angle D=180^\circ-60^\circ-70^\circ=50^\circ.
\displaystyle \angle B=\angle E=60^\circ,\qquad \angle C=\angle F=70^\circ.
\displaystyle BC=EF=4.5\text{ cm}.
\displaystyle \therefore\ \Delta ABC\cong\Delta DEF \quad\text{(ASA congruence criterion).}

\displaystyle \text{(ii) }DE=MN=6\text{ cm},\qquad EF=MR=8\text{ cm}.
\displaystyle \angle E=48^\circ\quad\text{and}\quad\angle R=48^\circ.
\displaystyle \text{However, }\angle E\text{ is the included angle between } DE\text{ and }EF,
\displaystyle \text{whereas }\angle R\text{ is not the included angle between } MN\text{ and }MR.
\displaystyle \text{Thus, the given information is SSA, which is not a}
\displaystyle \text{congruence criterion. Hence, the triangles cannot be}
\displaystyle \text{concluded to be congruent.}

\displaystyle \text{(iii) In }\Delta KLM,\quad \angle L=180^\circ-75^\circ-40^\circ=65^\circ.
\displaystyle \text{In }\Delta PQR,\quad \angle P=180^\circ-65^\circ-40^\circ=75^\circ.
\displaystyle \angle K=\angle P=75^\circ,\qquad \angle M=\angle R=40^\circ.
\displaystyle KM=PR=4\text{ cm}.
\displaystyle \therefore\ \Delta KLM\cong\Delta PQR \quad\text{(ASA congruence criterion).}

\displaystyle \text{(iv) In }\Delta ABC,\quad AC^2=BC^2-AB^2=5^2-3^2=16.
\displaystyle \therefore\ AC=4\text{ cm}.
\displaystyle \text{In }\Delta KLM,\quad KL^2=LM^2-KM^2=5^2-3^2=16.
\displaystyle \therefore\ KL=4\text{ cm}.
\displaystyle AB=KM=3\text{ cm},\qquad AC=KL=4\text{ cm},
\displaystyle \angle A=\angle K=90^\circ.
\displaystyle \therefore\ \Delta ABC\cong\Delta KML \quad\text{(SAS congruence criterion).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the adjoining figure, }P\text{ is the midpoint of }AB
\displaystyle \text{and }\angle PAC=\angle PBD.\text{ Prove that } \Delta PAC\cong\Delta PBD. p4\displaystyle \text{Answer:}
\displaystyle \angle PAC=\angle PBD\qquad\text{(Given)}.
\displaystyle AP=PB\qquad\text{(Since }P\text{ is the midpoint of }AB\text{)}.
\displaystyle \angle APC=\angle BPD\qquad\text{(Vertically opposite angles)}.
\displaystyle \therefore\ \Delta PAC\cong\Delta PBD \qquad\text{(ASA congruence criterion)}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In the adjoining figure, }DC\parallel AB \text{ and }\angle B=\angle D.
\displaystyle \text{Prove that }\Delta ABC\cong\Delta ADC. p5\displaystyle \text{Answer:}
\displaystyle \angle ABC=\angle CDA\qquad\text{(Given)}.
\displaystyle \angle BAC=\angle DCA \qquad\text{(Alternate interior angles, since }AB\parallel DC\text{)}.
\displaystyle AC=AC\qquad\text{(Common side)}.
\displaystyle \therefore\ \Delta ABC\cong\Delta CDA \qquad\text{(AAS congruence criterion)}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the adjoining figure, }BC=AD, \ \angle CAB=\angle ABD
\displaystyle \text{and }\angle ACB=\angle BDA.\text{ Prove that } \Delta ABC\cong\Delta BAD. p5\displaystyle \text{Answer:}
\displaystyle \text{In }\Delta ABC\text{ and }\Delta BAD,
\displaystyle \angle CAB=\angle ABD\qquad\text{(Given)}.
\displaystyle \angle ACB=\angle BDA\qquad\text{(Given)}.
\displaystyle BC=AD\qquad\text{(Given)}.
\displaystyle \therefore\ \Delta ABC\cong\Delta BAD \qquad\text{(AAS congruence criterion)}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the adjoining figure, }ABC\text{ is a triangle in which}
\displaystyle AB=AC,\ BL\perp AC\text{ and }CM\perp AB. \text{ Prove that }BL=CM. p6\displaystyle \text{Answer:}
\displaystyle \text{In }\Delta ABL\text{ and }\Delta ACM,
\displaystyle AB=AC\qquad\text{(Given)}.
\displaystyle \angle ALB=\angle AMC=90^\circ \qquad\text{(Since }BL\perp AC\text{ and }CM\perp AB\text{)}.
\displaystyle \angle BAL=\angle CAM\qquad\text{(Common angle at }A\text{)}.
\displaystyle \therefore\ \Delta ABL\cong\Delta ACM \qquad\text{(AAS congruence criterion)}.
\displaystyle \therefore\ BL=CM\qquad\text{(CPCTC)}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In }\Delta ABC,\ AB=AC. \text{ If }D\text{ is the midpoint of }BC,
\displaystyle \text{prove that: (i) }AD\text{ bisects }\angle A \text{ and (ii) }AD\perp BC.

\displaystyle \text{Answer:} screen-shot-2017-01-07-at-5-28-52-pm
\displaystyle \text{In }\Delta ABD\text{ and }\Delta ACD,
\displaystyle AB=AC\qquad\text{(Given)}.
\displaystyle BD=DC\qquad\text{(Since }D\text{ is the midpoint of }BC\text{)}.
\displaystyle AD=AD\qquad\text{(Common side)}.
\displaystyle \therefore\ \Delta ABD\cong\Delta ACD \qquad\text{(SSS congruence criterion)}.
\displaystyle \text{(i) }\angle BAD=\angle DAC\qquad\text{(CPCTC)}.
\displaystyle \therefore\ AD\text{ bisects }\angle BAC.
\displaystyle \text{(ii) }\angle ADB=\angle ADC\qquad\text{(CPCTC)}.
\displaystyle \angle ADB+\angle ADC=180^\circ \qquad\text{(Angles on a straight line)}.
\displaystyle \therefore\ \angle ADB=\angle ADC=90^\circ.
\displaystyle \therefore\ AD\perp BC.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In }\Delta ABC,\text{ it is given that }AB=AC \text{ and }AD\text{ is the}
\displaystyle \text{bisector of }\angle A,\text{ meeting }BC\text{ at }D. \text{ Prove that:}
\displaystyle \text{(i) }\Delta ABD\cong\Delta ACD \qquad\text{(ii) }AD\perp BC. p8
\displaystyle \text{Answer:}
\displaystyle AB=AC\qquad\text{(Given)}.
\displaystyle \angle ABD=\angle ACD \qquad\text{(Angles opposite equal sides are equal)}.
\displaystyle \angle BAD=\angle DAC \qquad\text{(Since }AD\text{ bisects }\angle A\text{)}.
\displaystyle \therefore\ \Delta ABD\cong\Delta ACD \qquad\text{(ASA congruence criterion)}.
\displaystyle \therefore\ \angle ADB=\angle ADC\qquad\text{(CPCTC)}.
\displaystyle \angle ADB+\angle ADC=180^\circ \qquad\text{(Angles on a straight line)}.
\displaystyle \therefore\ \angle ADB=\angle ADC=90^\circ.
\displaystyle \therefore\ AD\perp BC.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the adjoining figure, }\Delta ABC \text{ is such that }AB=AC
\displaystyle \text{and }\angle OBC=\angle OCB.\text{ Prove that:}
\displaystyle \text{(i) }\Delta ABO\cong\Delta ACO \qquad\text{(ii) }AO\text{ bisects }\angle A. p9\displaystyle \text{Answer:}
\displaystyle AB=AC\qquad\text{(Given)}.
\displaystyle \therefore\ \angle ABC=\angle ACB \qquad\text{(Angles opposite equal sides are equal)}.
\displaystyle \angle ABC=\angle ABO+\angle OBC.
\displaystyle \angle ACB=\angle ACO+\angle OCB.
\displaystyle \text{Also, }\angle OBC=\angle OCB\qquad\text{(Given)}.
\displaystyle \therefore\ \angle ABO=\angle ACO.
\displaystyle AO=AO\qquad\text{(Common side)}.
\displaystyle \therefore\ \Delta ABO\cong\Delta ACO \qquad\text{(AAS congruence criterion)}.
\displaystyle \therefore\ \angle BAO=\angle CAO\qquad\text{(CPCTC)}.
\displaystyle \therefore\ AO\text{ bisects }\angle BAC.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the adjoining figure, }AB\parallel GF,\ AC\parallel DE \text{ and }BF=CE.
\displaystyle \text{Prove that }\Delta BDE\cong\Delta FGC. p10\displaystyle \text{Answer:}
\displaystyle \text{In }\Delta BDE\text{ and }\Delta FGC,
\displaystyle \angle DBE=\angle GFC \qquad\text{(Corresponding angles, since }AB\parallel GF\text{)}.
\displaystyle \angle DEB=\angle GCF \qquad\text{(Corresponding angles, since }DE\parallel AC\text{)}.
\displaystyle BF=CE\qquad\text{(Given)}.
\displaystyle BF+FE=CE+FE.
\displaystyle \therefore\ BE=FC.
\displaystyle \therefore\ \Delta BDE\cong\Delta FGC \qquad\text{(ASA congruence criterion)}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the adjoining figure, }ABCD\text{ is a square and }CEB \text{ is an}
\displaystyle \text{isosceles triangle in which }EC=EB. \text{ Show that:}
\displaystyle \text{(i) }\Delta DCE\cong\Delta ABE \qquad\text{(ii) }AE=DE. p11\displaystyle \text{Answer:}
\displaystyle \text{In }\Delta DCE\text{ and }\Delta ABE,
\displaystyle DC=AB\qquad\text{(Sides of a square)}.
\displaystyle CE=BE\qquad\text{(Given)}.
\displaystyle \angle ECB=\angle CBE \qquad\text{(Base angles of isosceles }\Delta CEB\text{)}.
\displaystyle \angle DCB=\angle ABC=90^\circ \qquad\text{(Angles of a square)}.
\displaystyle \angle DCE=90^\circ+\angle BCE.
\displaystyle \angle ABE=90^\circ+\angle CBE.
\displaystyle \therefore\ \angle DCE=\angle ABE.
\displaystyle \therefore\ \Delta DCE\cong\Delta ABE \qquad\text{(SAS congruence criterion)}.
\displaystyle \therefore\ AE=DE\qquad\text{(CPCTC)}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the values of }x\text{ and }y \text{ in each of the following cases:}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }PN\parallel LM,\ PQ=KM. p12-1
\displaystyle \angle PQN=\angle MKL \qquad\text{(Vertically opposite angles)}.
\displaystyle \angle PNQ=\angle KLM \qquad\text{(Alternate interior angles, since }PN\parallel LM\text{)}.
\displaystyle \therefore\ \Delta PQN\cong\Delta MKL \qquad\text{(AAS congruence criterion)}.
\displaystyle \therefore\ PN=ML\qquad\text{(CPCTC)}.
\displaystyle 19=3x+4.
\displaystyle \therefore\ x=5.
\displaystyle \angle P=\angle M\qquad\text{(CPCTC)}.
\displaystyle 55^\circ=(2y-x)^\circ.
\displaystyle 55=2y-5.
\displaystyle \therefore\ y=30.

\displaystyle \text{(ii) In }\Delta ABC\text{ and }\Delta ADC, p12-2
\displaystyle AB=AD,\quad BC=DC\qquad\text{(Given)}.
\displaystyle AC=AC\qquad\text{(Common side)}.
\displaystyle \therefore\ \Delta ABC\cong\Delta ADC \qquad\text{(SSS congruence criterion)}.
\displaystyle \therefore\ \angle ABC=\angle ADC=103^\circ \qquad\text{(CPCTC)}.
\displaystyle 2x-5=103.
\displaystyle \therefore\ x=54.
\displaystyle \angle BAC=\angle CAD=34^\circ\qquad\text{(CPCTC)}.
\displaystyle 34^\circ+103^\circ+(7y+1)^\circ=180^\circ.
\displaystyle 7y+1=43.
\displaystyle \therefore\ y=6.

\displaystyle \text{(iii) In }\Delta PQS\text{ and }\Delta QPR, screen-shot-2017-01-07-at-5-53-55-pm
\displaystyle PS=QR,\quad QS=PR\qquad\text{(Given)}.
\displaystyle PQ=QP\qquad\text{(Common side)}.
\displaystyle \therefore\ \Delta PQS\cong\Delta QPR \qquad\text{(SSS congruence criterion)}.
\displaystyle \therefore\ \angle PSQ=\angle QRP=71^\circ \qquad\text{(CPCTC)}.
\displaystyle 2y-5=71.
\displaystyle \therefore\ y=38.
\displaystyle \angle PQS=\angle QPR=34^\circ \qquad\text{(CPCTC)}.
\displaystyle 3x+3=34.
\displaystyle \therefore\ x=\frac{31}{3}.

\displaystyle \text{(iv) From the figure, }RP=RQ. p12
\displaystyle \therefore\ \angle RPQ=\angle PQR \qquad\text{(Angles opposite equal sides are equal)}.
\displaystyle \therefore\ \angle RPM=\angle RQN \qquad\text{(Supplementary angles)}.
\displaystyle \text{However, no further given information in the figure establishes}
\displaystyle \angle MRP=\angle QRN\text{ or another condition sufficient to prove}
\displaystyle \Delta RMP\cong\Delta RNQ.
\displaystyle \text{Therefore, }x\text{ and }y\text{ cannot be uniquely determined from}
\displaystyle \text{the information shown in part (iv).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{In the adjoining figure, the sides }BA \text{ and }CA\text{ have been produced to }D\text{ and }E
\displaystyle \text{respectively, such that }BA=AD\text{ and }CA=AE. \text{ Prove that }ED\parallel BC. p13\displaystyle \text{Answer:}
\displaystyle \text{In }\Delta AED\text{ and }\Delta ACB,
\displaystyle AE=AC\qquad\text{(Given)}.
\displaystyle AD=AB\qquad\text{(Given)}.
\displaystyle \angle EAD=\angle CAB \qquad\text{(Vertically opposite angles)}.
\displaystyle \therefore\ \Delta AED\cong\Delta ACB \qquad\text{(SAS congruence criterion)}.
\displaystyle \therefore\ \angle ADE=\angle ABC \qquad\text{(CPCTC)}.
\displaystyle \text{Since }AD\text{ and }AB\text{ lie on the same straight line,}
\displaystyle \angle ADE=\angle ABC\text{ are alternate interior angles}.
\displaystyle \therefore\ ED\parallel BC.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Equilateral triangles }ABD\text{ and }ACE \text{ have been drawn on the sides }AB\text{ and }AC
\displaystyle \text{respectively of }\Delta ABC,\text{ as shown in the adjoining figure. Prove:}
\displaystyle \text{(i) }\angle DAC=\angle EAB \qquad \text{(ii) }DC=BE. p14\displaystyle \text{Answer:}
\displaystyle \text{Since }ABD\text{ and }ACE\text{ are equilateral triangles,}
\displaystyle AD=AB,\qquad AC=AE,
\displaystyle \angle DAB=\angle CAE=60^\circ.
\displaystyle \angle DAC=\angle DAB+\angle BAC =60^\circ+\angle BAC.
\displaystyle \angle EAB=\angle EAC+\angle CAB =60^\circ+\angle BAC.
\displaystyle \therefore\ \angle DAC=\angle EAB.
\displaystyle \text{In }\Delta DAC\text{ and }\Delta BAE,
\displaystyle AD=AB,\qquad AC=AE,\qquad \angle DAC=\angle BAE.
\displaystyle \therefore\ \Delta DAC\cong\Delta BAE \qquad\text{(SAS congruence criterion)}.
\displaystyle \therefore\ DC=BE\qquad\text{(CPCTC)}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In a regular pentagon }ABCDE, \text{ prove that }\Delta ABD\text{ is isosceles.} p15\displaystyle \text{Answer:}
\displaystyle \text{Since }ABCDE\text{ is a regular pentagon, all its sides and all its interior angles are equal.}
\displaystyle \therefore\ AE=BC,\qquad ED=CD,
\displaystyle \angle AED=\angle BCD=108^\circ.
\displaystyle \text{In }\Delta AED\text{ and }\Delta BCD,
\displaystyle AE=BC,\qquad ED=CD,\qquad \angle AED=\angle BCD.
\displaystyle \therefore\ \Delta AED\cong\Delta BCD \qquad\text{(SAS congruence criterion)}.
\displaystyle \therefore\ AD=BD\qquad\text{(CPCTC)}.
\displaystyle \text{Hence, }\Delta ABD\text{ is an isosceles triangle.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoining figure, }ABCD \text{ is a square, and }P,Q,R\text{ are points on}
\displaystyle AB,BC,CD\text{ respectively, such that }AP=BQ=CR \text{ and }\angle PQR=90^\circ.
\displaystyle \text{Prove: (i) }PB=QC\qquad\text{(ii) }PQ=QR \qquad\text{(iii) }\angle QPR=45^\circ. p16\displaystyle \text{Answer:}
\displaystyle \text{Since }ABCD\text{ is a square, }AB=BC.
\displaystyle AP+PB=BQ+QC.
\displaystyle \text{But }AP=BQ\qquad\text{(Given)}.
\displaystyle \therefore\ PB=QC.
\displaystyle \text{Hence, (i) }PB=QC.
\displaystyle \text{Now, in }\Delta PBQ\text{ and }\Delta QCR,
\displaystyle PB=QC\qquad\text{(Proved above)},
\displaystyle BQ=CR\qquad\text{(Given)},
\displaystyle \angle PBQ=\angle QCR=90^\circ \qquad\text{(Angles of a square)}.
\displaystyle \therefore\ \Delta PBQ\cong\Delta QCR \qquad\text{(SAS congruence criterion)}.
\displaystyle \therefore\ PQ=QR\qquad\text{(CPCTC)}.
\displaystyle \text{Hence, (ii) }PQ=QR.
\displaystyle \text{Therefore, }\Delta PQR\text{ is an isosceles triangle.}
\displaystyle \text{Also, }\angle PQR=90^\circ\qquad\text{(Given)}.
\displaystyle \angle QPR=\angle PRQ =\frac{180^\circ-90^\circ}{2}=45^\circ.
\displaystyle \text{Hence, (iii) }\angle QPR=45^\circ.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, }QK\parallel ML, \ QM\parallel KL\text{ and }RL=LP.
\displaystyle \text{Prove: (i) }\angle MPL=\angle KLR \qquad\text{(ii) }ML=QK. p18\displaystyle \text{Answer:}
\displaystyle \text{Since }ML\parallel QK\text{ and }P,M,Q \text{ are collinear,}
\displaystyle \angle PML=\angle KQL \qquad\text{(Corresponding angles)}.
\displaystyle \text{Since }KL\parallel QM\text{ and }P,L,R \text{ are collinear,}
\displaystyle \angle PLM=\angle KRQ \qquad\text{(Corresponding angles)}.
\displaystyle \therefore\ \angle MPL=\angle QKL.
\displaystyle \text{Since }QK\parallel ML,\quad \angle QKL=\angle KLM\qquad\text{(Alternate angles)}.
\displaystyle \therefore\ \angle MPL=\angle KLM.
\displaystyle \text{Since }P,L,R\text{ are collinear, } \angle KLM+\angle KLR=180^\circ.
\displaystyle \text{Thus the proposed statement } \angle MPL=\angle KLR\text{ does not follow from the figure.}
\displaystyle \text{In fact, }\angle MPL+\angle KLR=180^\circ.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the adjoining figure, }ABCD \text{ is a square, }EF\parallel AC\text{ and }R
\displaystyle \text{is the midpoint of }EF.\text{ Prove: (i) }AE=CF \quad\text{(ii) }DE=DF
\displaystyle \text{(iii) }DR\text{ bisects }\angle EDF. p17\displaystyle \text{Answer:}
\displaystyle \text{Since }EF\parallel AC,\quad \angle BEF=\angle BAC=45^\circ.
\displaystyle \text{Also, }\angle EBF=90^\circ \qquad\text{(Angle of a square)}.
\displaystyle \therefore\ \angle BFE=180^\circ-90^\circ-45^\circ=45^\circ.
\displaystyle \therefore\ BE=BF \qquad\text{(Sides opposite equal angles are equal)}.
\displaystyle \text{Also, }AB=BC\qquad\text{(Sides of a square)}.
\displaystyle AE=AB-BE\quad\text{and}\quad CF=BC-BF.
\displaystyle \therefore\ AE=CF.
\displaystyle \text{Hence, (i) }AE=CF.
\displaystyle \text{Now, in }\Delta DAE\text{ and }\Delta DCF,
\displaystyle DA=DC\qquad\text{(Sides of a square)},
\displaystyle AE=CF\qquad\text{(Proved above)},
\displaystyle \angle DAE=\angle DCF=90^\circ.
\displaystyle \therefore\ \Delta DAE\cong\Delta DCF \qquad\text{(SAS congruence criterion)}.
\displaystyle \therefore\ DE=DF\qquad\text{(CPCTC)}.
\displaystyle \text{Hence, (ii) }DE=DF.
\displaystyle \text{Since }R\text{ is the midpoint of }EF,\quad ER=RF.
\displaystyle \text{In }\Delta DER\text{ and }\Delta DFR,
\displaystyle DE=DF,\qquad ER=RF,\qquad DR=DR\text{ (common)}.
\displaystyle \therefore\ \Delta DER\cong\Delta DFR \qquad\text{(SSS congruence criterion)}.
\displaystyle \therefore\ \angle EDR=\angle RDF\qquad\text{(CPCTC)}.
\displaystyle \therefore\ DR\text{ bisects }\angle EDF.
\displaystyle \text{Hence, (iii) is proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In the adjoining figure, } \angle TPQ=\angle SQP,\ \angle SRP=\angle TRQ
\displaystyle \text{and }R\text{ is the midpoint of }PQ. \text{ Prove: (i) }\Delta PRT\cong\Delta QRS
\displaystyle \text{(ii) }PT=SQ\qquad \text{(iii) }\angle PTR=\angle QSR. screen-shot-2017-01-07-at-5-23-41-pm\displaystyle \text{Answer:}
\displaystyle \text{Since }R\text{ is the midpoint of }PQ,\quad PR=RQ.
\displaystyle \angle TPR=\angle SQR \qquad\text{(Given, since }P,R,Q\text{ are collinear)}.
\displaystyle \angle TRQ=\angle SRP\qquad\text{(Given)}.
\displaystyle \text{Since }P,R,Q\text{ are collinear,}
\displaystyle \angle PRT+\angle TRQ=180^\circ,
\displaystyle \angle QRS+\angle SRP=180^\circ.
\displaystyle \text{Since }\angle TRQ=\angle SRP,\quad \angle PRT=\angle QRS.
\displaystyle \text{Thus, in }\Delta PRT\text{ and }\Delta QRS,
\displaystyle \angle TPR=\angle SQR,\quad PR=RQ,\quad \angle PRT=\angle QRS.
\displaystyle \therefore\ \Delta PRT\cong\Delta QRS \qquad\text{(ASA congruence criterion)}.
\displaystyle \text{Hence, (i) is proved.}
\displaystyle \therefore\ PT=SQ\qquad\text{(CPCTC)}.
\displaystyle \text{Hence, (ii) is proved.}
\displaystyle \therefore\ \angle PTR=\angle QSR \qquad\text{(CPCTC)}.
\displaystyle \text{Hence, (iii) is proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the adjoining figure, } \angle PON=90^\circ\text{ and }MO=ON.
\displaystyle \text{Prove: (i) }PM=PN\qquad \text{(ii) }\angle OMQ=\angle ONQ. screen-shot-2017-01-07-at-5-27-10-pm\displaystyle \text{Answer:}
\displaystyle \text{Since }M,O,N\text{ are collinear and } \angle PON=90^\circ,
\displaystyle \angle POM=\angle PON=90^\circ.
\displaystyle \text{In }\Delta POM\text{ and }\Delta PON,
\displaystyle MO=ON\qquad\text{(Given)},
\displaystyle PO=PO\qquad\text{(Common side)},
\displaystyle \angle POM=\angle PON=90^\circ.
\displaystyle \therefore\ \Delta POM\cong\Delta PON \qquad\text{(SAS congruence criterion)}.
\displaystyle \therefore\ PM=PN\qquad\text{(CPCTC)}.
\displaystyle \text{Hence, (i) is proved.}
\displaystyle \text{Since }P,O,Q\text{ are collinear and } \angle PON=90^\circ,
\displaystyle \angle MOQ=\angle NOQ=90^\circ.
\displaystyle \text{Now, in }\Delta MOQ\text{ and }\Delta NOQ,
\displaystyle MO=ON\qquad\text{(Given)},
\displaystyle OQ=OQ\qquad\text{(Common side)},
\displaystyle \angle MOQ=\angle NOQ=90^\circ.
\displaystyle \therefore\ \Delta MOQ\cong\Delta NOQ \qquad\text{(SAS congruence criterion)}.
\displaystyle \therefore\ \angle OMQ=\angle ONQ \qquad\text{(CPCTC)}.
\displaystyle \text{Hence, (ii) is proved.}
\displaystyle \\

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