\displaystyle \textbf{Question 1: } \text{State True or False:}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Two similar polygons are necessarily congruent. } \text{False}
\displaystyle \text{(ii) Two congruent polygons are necessarily similar. } \text{True}
\displaystyle \text{(iii) All equiangular triangles are similar. } \text{True}
\displaystyle \text{(iv) All isosceles triangles are similar. } \text{False}
\displaystyle \text{(v) Two isosceles right-angled triangles are similar. } \text{True}
\displaystyle \text{(vi) Two isosceles triangles are similar if an angle of one is congruent}
\displaystyle \text{to the corresponding angle of the other. } \text{True}
\displaystyle \text{(vii) The diagonals of a trapezium divide each other into proportional}
\displaystyle \text{segments. } \text{True}
\\

\displaystyle \textbf{Question 2: } \text{In }\triangle ABC,\ DE\parallel BC,\text{ where }D\text{ and }E\text{ are points}
\displaystyle \text{on }AB\text{ and }AC\text{ respectively. Prove that }\triangle ADE\sim\triangle ABC.
\displaystyle \text{Also find the length of }DE,\text{ if }AD=12\text{ cm},BD=24\text{ cm and }BC=8\text{ cm}.
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,\ DE\parallel BC
\displaystyle \text{Consider }\triangle ABC\text{ and }\triangle ADE.
\displaystyle \angle ABC=\angle ADE\quad\text{(corresponding angles)}
\displaystyle \angle ACB=\angle AED\quad\text{(corresponding angles)}
\displaystyle \therefore \triangle ADE\sim\triangle ABC\quad\text{(AA similarity)}
\displaystyle \text{Now, corresponding sides are proportional.}
\displaystyle \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}
\displaystyle AB=AD+BD=12+24=36\text{ cm}
\displaystyle \frac{AD}{AB}=\frac{DE}{BC}
\displaystyle \frac{12}{36}=\frac{DE}{8}
\displaystyle DE=\frac{12}{36}\times8=\frac{8}{3}\text{ cm}
\displaystyle \therefore DE=\frac{8}{3}\text{ cm}
\\

\displaystyle \textbf{Question 3: } \text{Given }\angle GHE=\angle DFE=90^\circ,
\displaystyle DH=8,\ DF=12,\ DG=3x-1\text{ and }DE=4x+2.
\displaystyle \text{Find the length of the segments }DG\text{ and }DE.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle EFD\text{ and }\triangle GHD.
\displaystyle \angle DFE=\angle GHD=90^\circ\quad\text{(given)}
\displaystyle \angle FDE=\angle HDG\quad\text{(common angle at }D\text{)}
\displaystyle \therefore \triangle EFD\sim\triangle GHD\quad\text{(AA similarity)}
\displaystyle \text{Hence, corresponding sides are proportional.}
\displaystyle \frac{DH}{DF}=\frac{DG}{DE}
\displaystyle \frac{8}{12}=\frac{3x-1}{4x+2}
\displaystyle 8(4x+2)=12(3x-1)
\displaystyle 32x+16=36x-12
\displaystyle 4x=28
\displaystyle x=7
\displaystyle DG=3x-1=3(7)-1=20
\displaystyle DE=4x+2=4(7)+2=30
\displaystyle \therefore DG=20\text{ and }DE=30
\\

\displaystyle \textbf{Question 4: } D\text{ is a point of side }BC\text{ of }\triangle ABC\text{ such that}
\displaystyle \angle ADC=\angle BAC.\text{ Prove that }CA^2=CB\times CD.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle ABC\text{ and }\triangle ADC.
\displaystyle \angle BAC=\angle ADC\quad\text{(given)}
\displaystyle \angle ACB=\angle ACD\quad\text{(common angle at }C\text{)}
\displaystyle \therefore \triangle ABC\sim\triangle ADC\quad\text{(AA similarity)}
\displaystyle \text{Hence, corresponding sides are proportional.}
\displaystyle \frac{CA}{CB}=\frac{CD}{CA}
\displaystyle CA\times CA=CB\times CD
\displaystyle \therefore CA^2=CB\times CD
\\

\displaystyle \textbf{Question 5: } \text{In the given figure, }\triangle ABC\text{ and }\triangle AMP\text{ are}
\displaystyle \text{right angled at }B\text{ and }M\text{ respectively. Given }AC=10\text{ cm},
\displaystyle AP=15\text{ cm and }PM=12\text{ cm}.
\displaystyle \text{(i) Prove }\triangle ABC\sim\triangle AMP\qquad \text{(ii) Find }AB\text{ and }BC\quad\text{[ICSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC\text{ and }\triangle AMP,
\displaystyle \angle ABC=\angle AMP=90^\circ\quad\text{(given)}
\displaystyle \angle BAC=\angle PAM\quad\text{(common angle)}
\displaystyle \therefore \triangle ABC\sim\triangle AMP\quad\text{(AA similarity)}
\displaystyle \text{Now, in right }\triangle AMP,
\displaystyle AP^2=AM^2+PM^2
\displaystyle 15^2=AM^2+12^2
\displaystyle AM^2=225-144=81
\displaystyle AM=9\text{ cm}
\displaystyle \text{Since }\triangle ABC\sim\triangle AMP,
\displaystyle \frac{AB}{AM}=\frac{AC}{AP}=\frac{BC}{PM}
\displaystyle \frac{AB}{9}=\frac{10}{15}
\displaystyle AB=9\times\frac{10}{15}=6\text{ cm}
\displaystyle \frac{BC}{12}=\frac{10}{15}
\displaystyle BC=12\times\frac{10}{15}=8\text{ cm}
\displaystyle \therefore AB=6\text{ cm and }BC=8\text{ cm}
\\

\displaystyle \textbf{Question 6: } E\text{ and }F\text{ are points on sides }DC\text{ and }AB\text{ respectively}
\displaystyle \text{of parallelogram }ABCD.\text{ If diagonal }AC\text{ and segment }EF\text{ intersect at }G,
\displaystyle \text{prove that }AG\times EG=FG\times CG.
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle AGF\text{ and }\triangle EGC,
\displaystyle \angle AGF=\angle EGC\quad\text{(vertically opposite angles)}
\displaystyle \angle FAG=\angle ECG\quad\text{(alternate angles, since }AB\parallel DC\text{)}
\displaystyle \therefore \triangle AGF\sim\triangle CGE\quad\text{(AA similarity)}
\displaystyle \text{Hence, corresponding sides are proportional.}
\displaystyle \frac{AG}{CG}=\frac{FG}{EG}
\displaystyle AG\times EG=FG\times CG
\displaystyle \therefore AG\times EG=FG\times CG
\\

\displaystyle \textbf{Question 7: } \text{Given }RS\text{ and }PT\text{ are altitudes of }\triangle PQR.
\displaystyle \text{Prove that (i) }\triangle PQT\sim\triangle QRS\qquad\text{(ii) }PQ\times QS=RQ\times QT.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle PQT\text{ and }\triangle QRS.
\displaystyle \angle QTP=\angle QSR=90^\circ\quad\text{(given, altitudes)}
\displaystyle \angle PQT=\angle SQR\quad\text{(common angle at }Q\text{)}
\displaystyle \therefore \triangle PQT\sim\triangle QRS\quad\text{(AA similarity)}
\displaystyle \text{(ii) Since }\triangle PQT\sim\triangle QRS,
\displaystyle \frac{PQ}{RQ}=\frac{QT}{QS}
\displaystyle PQ\times QS=RQ\times QT
\displaystyle \therefore PQ\times QS=RQ\times QT
\\

\displaystyle \textbf{Question 8: } \text{Given }ABCD\text{ is a rhombus, }DPR\text{ and }CBR\text{ are}
\displaystyle \text{straight lines. Prove that }DP\times CR=DC\times PR.
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle DPA\text{ and }\triangle RPC,
\displaystyle \angle DPA=\angle RPC\quad\text{(vertically opposite angles)}
\displaystyle \angle PAD=\angle PCR\quad\text{(alternate angles, since }AD\parallel CR\text{)}
\displaystyle \therefore \triangle DPA\sim\triangle RPC\quad\text{(AA similarity)}
\displaystyle \text{Hence, corresponding sides are proportional.}
\displaystyle \frac{DP}{PR}=\frac{AD}{CR}
\displaystyle \text{But }AD=DC\quad\text{(sides of a rhombus)}
\displaystyle \therefore \frac{DP}{PR}=\frac{DC}{CR}
\displaystyle DP\times CR=DC\times PR
\displaystyle \therefore DP\times CR=DC\times PR
\\

\displaystyle \textbf{Question 9: } \text{Given }FB=FD,\ AE\perp FD\text{ and }FC\perp AD.
\displaystyle \text{Prove that }\frac{FB}{AD}=\frac{BC}{ED}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }FB=FD,
\displaystyle \angle FDB=\angle FBD\quad\text{(angles opposite equal sides)}
\displaystyle \text{Consider }\triangle AED\text{ and }\triangle FCB.
\displaystyle \angle AED=\angle FCB=90^\circ\quad\text{(given)}
\displaystyle \angle ADE=\angle FBC\quad\text{(since }\angle FDB=\angle FBD\text{)}
\displaystyle \therefore \triangle AED\sim\triangle FCB\quad\text{(AA similarity)}
\displaystyle \text{Hence, corresponding sides are proportional.}
\displaystyle \frac{AD}{FB}=\frac{ED}{BC}
\displaystyle \therefore \frac{FB}{AD}=\frac{BC}{ED}
\\

\displaystyle \textbf{Question 10: } \text{In }\triangle ABC,\angle B=2\angle C\text{ and the bisector of }\angle B
\displaystyle \text{meets }CA\text{ at point }D.\text{ Prove that (i) }\triangle ABC\sim\triangle ABD
\displaystyle \text{(ii) }DC:AD=BC:AB.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle B=2\angle C
\displaystyle \text{Since }BD\text{ bisects }\angle B,
\displaystyle \angle ABD=\angle DBC=\frac{1}{2}\angle B=\angle C
\displaystyle \therefore \angle ABD=\angle ACB
\displaystyle \text{Also, }\angle BAC=\angle DAB\quad\text{(common angle, since }D\text{ lies on }CA\text{)}
\displaystyle \therefore \triangle ABC\sim\triangle ABD\quad\text{(AA similarity)}
\displaystyle \text{From }\triangle ABC\sim\triangle ABD,
\displaystyle \frac{BC}{BD}=\frac{AB}{AD}
\displaystyle \text{Now, }\angle DBC=\angle DCB
\displaystyle \therefore BD=DC\quad\text{(sides opposite equal angles)}
\displaystyle \frac{BC}{DC}=\frac{AB}{AD}
\displaystyle BC\times AD=AB\times DC
\displaystyle \frac{DC}{AD}=\frac{BC}{AB}
\displaystyle \therefore DC:AD=BC:AB
\\

\displaystyle \textbf{Question 11: } \text{In }\triangle PQR,\angle Q=90^\circ\text{ and }QM\perp PR.
\displaystyle \text{Prove that:}
\displaystyle \text{(i) }PQ^2=PM\times PR
\displaystyle \text{(ii) }QR^2=PR\times MR
\displaystyle \text{(iii) }PQ^2+QR^2=PR^2
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle PQM\text{ and }\triangle PRQ.
\displaystyle \angle PMQ=\angle PQR=90^\circ\quad\text{(given)}
\displaystyle \angle QPM=\angle RPQ\quad\text{(common angle)}
\displaystyle \therefore \triangle PQM\sim\triangle PRQ\quad\text{(AA similarity)}
\displaystyle \text{Hence, }\frac{PQ}{PR}=\frac{PM}{PQ}
\displaystyle \therefore PQ^2=PM\times PR
\displaystyle \text{(ii) Consider }\triangle QMR\text{ and }\triangle PQR.
\displaystyle \angle QMR=\angle PQR=90^\circ\quad\text{(given)}
\displaystyle \angle QRM=\angle QRP\quad\text{(common angle)}
\displaystyle \therefore \triangle QMR\sim\triangle PQR\quad\text{(AA similarity)}
\displaystyle \text{Hence, }\frac{QR}{PR}=\frac{MR}{QR}
\displaystyle \therefore QR^2=PR\times MR
\displaystyle \text{(iii) Adding (i) and (ii),}
\displaystyle PQ^2+QR^2=PM\times PR+PR\times MR
\displaystyle PQ^2+QR^2=PR(PM+MR)
\displaystyle \text{Since }PM+MR=PR,
\displaystyle PQ^2+QR^2=PR\times PR=PR^2
\displaystyle \therefore PQ^2+QR^2=PR^2
\\

\displaystyle \textbf{Question 12: } \text{In }\triangle ABC,\angle C=90^\circ,\ CD\perp AB.
\displaystyle \text{Prove that }CD^2=AD\times DB.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle ACD\text{ and }\triangle CBD.
\displaystyle \angle ADC=\angle CDB=90^\circ\quad\text{(given)}
\displaystyle \angle ACD=\angle CBD\quad\text{(each is complementary to }\angle CAB\text{)}
\displaystyle \therefore \triangle ACD\sim\triangle CBD\quad\text{(AA similarity)}
\displaystyle \text{Hence, corresponding sides are proportional.}
\displaystyle \frac{AD}{CD}=\frac{CD}{DB}
\displaystyle CD\times CD=AD\times DB
\displaystyle \therefore CD^2=AD\times DB
\\

\displaystyle \textbf{Question 13: } \text{In }\triangle ABC,\angle B=90^\circ,\ BD\perp AC.
\displaystyle \text{(i) If }CD=10\text{ cm and }BD=8\text{ cm, find }AD.
\displaystyle \text{(ii) If }AC=18\text{ cm and }AD=6\text{ cm, find }BD.
\displaystyle \text{(iii) If }AC=9\text{ cm and }AB=7\text{ cm, find }AD.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle CDB\text{ and }\triangle BDA.
\displaystyle \angle CDB=\angle BDA=90^\circ\quad\text{(given)}
\displaystyle \angle CBD=\angle BAD\quad\text{(complementary to }\angle A\text{)}
\displaystyle \therefore \triangle CDB\sim\triangle BDA\quad\text{(AA similarity)}
\displaystyle \text{(i) }\frac{CD}{BD}=\frac{BD}{AD}
\displaystyle AD=\frac{BD^2}{CD}=\frac{8^2}{10}=6.4\text{ cm}
\displaystyle \text{(ii) }CD=AC-AD=18-6=12\text{ cm}
\displaystyle BD^2=AD\times CD=6\times12=72
\displaystyle BD=\sqrt{72}=6\sqrt{2}\approx8.49\text{ cm}
\displaystyle \text{(iii) From }\triangle ABC\sim\triangle ADB,
\displaystyle \frac{AD}{AB}=\frac{AB}{AC}
\displaystyle AD=\frac{AB^2}{AC}=\frac{7^2}{9}=\frac{49}{9}\text{ cm}
\displaystyle AD\approx5.44\text{ cm}
\\

\displaystyle \textbf{Question 14: } \text{In the figure, }PQRS\text{ is a parallelogram with }PQ=16\text{ cm}
\displaystyle \text{and }QR=10\text{ cm. }L\text{ is a point on }PR\text{ such that }RL:LP=2:3.
\displaystyle QL\text{ produced meets }RS\text{ at }M\text{ and }PS\text{ produced at }N.
\displaystyle \text{Find the lengths of }PN\text{ and }RM.\quad\text{[ICSE 1997]}
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle RLQ\text{ and }\triangle PLN,
\displaystyle \angle RLQ=\angle PLN\quad\text{(vertically opposite angles)}
\displaystyle \angle LRQ=\angle LPN\quad\text{(alternate angles, since }RQ\parallel PN\text{)}
\displaystyle \therefore \triangle RLQ\sim\triangle PLN\quad\text{(AA similarity)}
\displaystyle \text{Hence, }\frac{RL}{LP}=\frac{RQ}{PN}
\displaystyle \frac{2}{3}=\frac{10}{PN}
\displaystyle PN=\frac{10\times3}{2}=15\text{ cm}
\displaystyle \text{Now, in }\triangle RLM\text{ and }\triangle PLQ,
\displaystyle \angle RLM=\angle PLQ\quad\text{(vertically opposite angles)}
\displaystyle \angle LRM=\angle LPQ\quad\text{(alternate angles, since }RM\parallel PQ\text{)}
\displaystyle \therefore \triangle RLM\sim\triangle PLQ\quad\text{(AA similarity)}
\displaystyle \text{Hence, }\frac{RM}{PQ}=\frac{RL}{LP}
\displaystyle \frac{RM}{16}=\frac{2}{3}
\displaystyle RM=\frac{2}{3}\times16=\frac{32}{3}\text{ cm}
\displaystyle RM=10.67\text{ cm}
\displaystyle \therefore PN=15\text{ cm and }RM=\frac{32}{3}\text{ cm}
\\

\displaystyle \textbf{Question 15: } \text{In quadrilateral }ABCD,\text{ diagonals }AC\text{ and }BD\text{ intersect}
\displaystyle \text{at point }E\text{ such that }AE=EC\text{ and }BE=ED.\text{ Show that }ABCD
\displaystyle \text{is a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AE=EC\text{ and }BE=ED
\displaystyle \therefore E\text{ is the midpoint of }AC\text{ and }BD.
\displaystyle \text{Hence, the diagonals }AC\text{ and }BD\text{ bisect each other.}
\displaystyle \text{A quadrilateral whose diagonals bisect each other is a parallelogram.}
\displaystyle \therefore ABCD\text{ is a parallelogram.}
\\

\displaystyle \textbf{Question 16: } \text{Given }AB\parallel DE\text{ and }BC\parallel EF.\text{ Prove that:}
\displaystyle \text{(i) }\frac{AD}{DG}=\frac{CF}{FG}
\displaystyle \text{(ii) }\triangle DFG\sim\triangle ACG
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle AGB,\ DE\parallel AB
\displaystyle \therefore \frac{AD}{DG}=\frac{EB}{GE}\quad\text{(basic proportionality theorem)}
\displaystyle \text{In }\triangle BGC,\ EF\parallel BC
\displaystyle \therefore \frac{EB}{GE}=\frac{CF}{FG}\quad\text{(basic proportionality theorem)}
\displaystyle \therefore \frac{AD}{DG}=\frac{CF}{FG}
\displaystyle \text{This proves part (i).}
\displaystyle \text{Now, }\frac{AD}{DG}=\frac{CF}{FG}
\displaystyle \frac{AG-DG}{DG}=\frac{CG-FG}{FG}
\displaystyle \frac{AG}{DG}-1=\frac{CG}{FG}-1
\displaystyle \frac{AG}{DG}=\frac{CG}{FG}
\displaystyle \therefore \frac{DG}{AG}=\frac{FG}{CG}
\displaystyle \text{Also, }\angle DGF=\angle AGC\quad\text{(same angle at }G\text{)}
\displaystyle \therefore \triangle DFG\sim\triangle ACG\quad\text{(SAS similarity)}
\\

\displaystyle \textbf{Question 17: } \text{In }\triangle ABC,\ AD\perp BC\text{ and }AD^2=BD\times DC.
\displaystyle \text{Show that }\angle BAC=90^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AD^2=BD\times DC
\displaystyle \frac{AD}{DC}=\frac{BD}{AD}
\displaystyle \text{Also, }\angle ADB=\angle ADC=90^\circ\quad\text{(since }AD\perp BC\text{)}
\displaystyle \therefore \triangle DBA\sim\triangle DAC\quad\text{(SAS similarity)}
\displaystyle \therefore \angle C=\angle BAD\text{ and }\angle B=\angle DAC
\displaystyle \text{Adding, }\angle C+\angle B=\angle BAD+\angle DAC
\displaystyle \angle B+\angle C=\angle BAC\qquad\cdots(i)
\displaystyle \text{In }\triangle ABC,\ \angle A+\angle B+\angle C=180^\circ
\displaystyle \angle BAC+\angle B+\angle C=180^\circ\qquad\cdots(ii)
\displaystyle \text{From (i) and (ii), }\angle BAC+\angle BAC=180^\circ
\displaystyle 2\angle BAC=180^\circ
\displaystyle \therefore \angle BAC=90^\circ
\\

\displaystyle \textbf{Question 18: } \text{In the given figure }AB\parallel EF\parallel DC;
\displaystyle AB=67.5\text{ cm},\ DC=40.5\text{ cm and }AE=52.5\text{ cm}.
\displaystyle \text{(i) Name the three pairs of similar triangles.}
\displaystyle \text{(ii) Find the lengths of }EC\text{ and }EF.
\displaystyle \text{Answer:}
\displaystyle \text{The three pairs of similar triangles are:}
\displaystyle \triangle ABE\sim\triangle CDE
\displaystyle \triangle ABC\sim\triangle CEF
\displaystyle \triangle BCD\sim\triangle BEF
\displaystyle \text{Since }\triangle ABE\sim\triangle CDE,
\displaystyle \frac{AB}{CD}=\frac{AE}{CE}
\displaystyle \frac{67.5}{40.5}=\frac{52.5}{CE}
\displaystyle CE=\frac{52.5\times40.5}{67.5}=31.5\text{ cm}
\displaystyle AC=AE+CE=52.5+31.5=84\text{ cm}
\displaystyle \text{Since }\triangle ABC\sim\triangle CEF,
\displaystyle \frac{EF}{AB}=\frac{CE}{AC}
\displaystyle EF=67.5\times\frac{31.5}{84}
\displaystyle EF=25.3125\text{ cm}
\displaystyle \therefore EC=31.5\text{ cm and }EF=25.3125\text{ cm}
\\

\displaystyle \textbf{Question 19: } \text{In the given figure }QR\parallel AB\text{ and }DR\parallel QB.
\displaystyle \text{Prove that }PQ^2=PD\times PA.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }QR\parallel AB\text{ and }DR\parallel QB
\displaystyle \text{In }\triangle PAB,\ QR\parallel AB
\displaystyle \therefore \frac{PQ}{PA}=\frac{PR}{PB}\qquad\cdots(i)
\displaystyle \text{In }\triangle PQB,\ DR\parallel QB
\displaystyle \therefore \frac{PD}{PQ}=\frac{PR}{PB}\qquad\cdots(ii)
\displaystyle \text{From (i) and (ii),}
\displaystyle \frac{PQ}{PA}=\frac{PD}{PQ}
\displaystyle PQ\times PQ=PD\times PA
\displaystyle \therefore PQ^2=PD\times PA
\\

\displaystyle \textbf{Question 20: } \text{Through the midpoint }M\text{ of the side }CD\text{ of a parallelogram}
\displaystyle ABCD,\text{ the line }BM\text{ is drawn intersecting diagonal }AC\text{ in }L\text{ and }AD
\displaystyle \text{produced in }E.\text{ Prove that }EL=2BL.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle DEM\text{ and }\triangle CBM.
\displaystyle \angle EDM=\angle BCM\quad\text{(alternate angles, since }DE\parallel BC\text{)}
\displaystyle \angle DME=\angle CMB\quad\text{(vertically opposite angles)}
\displaystyle DM=MC\quad\text{(}M\text{ is the midpoint of }CD\text{)}
\displaystyle \therefore \triangle DEM\cong\triangle CBM\quad\text{(ASA congruence)}
\displaystyle \therefore DE=BC\quad\text{(corresponding sides)}
\displaystyle \text{Also, }AD=BC\quad\text{(opposite sides of a parallelogram)}
\displaystyle AE=AD+DE=BC+BC=2BC
\displaystyle \text{Now, consider }\triangle ELA\text{ and }\triangle BLC.
\displaystyle \angle ELA=\angle BLC\quad\text{(vertically opposite angles)}
\displaystyle \angle EAL=\angle BCL\quad\text{(alternate angles, since }AE\parallel BC\text{)}
\displaystyle \therefore \triangle ELA\sim\triangle BLC\quad\text{(AA similarity)}
\displaystyle \frac{EL}{BL}=\frac{EA}{BC}
\displaystyle \frac{EL}{BL}=\frac{2BC}{BC}=2
\displaystyle EL=2BL
\\

\displaystyle \textbf{Question 21: } \text{In the figure given below, }P\text{ is a point on }AB\text{ such that}
\displaystyle AP:PB=4:3.\ PQ\parallel AC.
\displaystyle \text{(i) Calculate the ratio }PQ:AC,\text{ giving reasons for your answer.}
\displaystyle \text{(ii) In }\triangle ARC,\angle ARC=90^\circ\text{ and in }\triangle PQS,\angle PSQ=90^\circ.
\displaystyle \text{Given }QS=6\text{ cm, calculate the length of }AR.\quad\text{[ICSE 1999]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }AP:PB=4:3
\displaystyle \text{Since }PQ\parallel AC,
\displaystyle \frac{AP}{PB}=\frac{CQ}{QB}\quad\text{(basic proportionality theorem)}
\displaystyle \frac{CQ}{QB}=\frac{4}{3}
\displaystyle \therefore \frac{BQ}{BC}=\frac{3}{7}
\displaystyle \text{Also, }\angle QBP=\angle ACB\quad\text{(corresponding angles)}
\displaystyle \angle QPB=\angle CAB\quad\text{(corresponding angles)}
\displaystyle \therefore \triangle PBQ\sim\triangle ABC\quad\text{(AA similarity)}
\displaystyle \therefore \frac{PQ}{AC}=\frac{BQ}{BC}=\frac{3}{7}
\displaystyle \therefore PQ:AC=3:7
\displaystyle \text{(ii) Given, }\angle ARC=\angle QSP=90^\circ
\displaystyle \angle ACR=\angle PQS\quad\text{(alternate angles, since }AC\parallel PQ\text{)}
\displaystyle \therefore \triangle ARC\sim\triangle QSP\quad\text{(AA similarity)}
\displaystyle \frac{AR}{QS}=\frac{AC}{PQ}
\displaystyle \frac{AR}{6}=\frac{7}{3}
\displaystyle AR=\frac{7\times6}{3}=14\text{ cm}
\displaystyle \therefore AR=14\text{ cm}
\\

\displaystyle \textbf{Question 22: } \text{In the right angled }\triangle QPR,\ PM\text{ is the altitude. Given that}
\displaystyle QR=8\text{ cm and }MQ=3.5\text{ cm, calculate the value of }PR.\quad\text{[ICSE 2000]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }QR=8\text{ cm and }MQ=3.5\text{ cm,}
\displaystyle MR=QR-MQ=8-3.5=4.5\text{ cm}
\displaystyle \text{In right angled }\triangle QPR,\ PM\perp QR
\displaystyle PR^2=QR\times MR\quad\text{(projection theorem)}
\displaystyle PR^2=8\times4.5=36
\displaystyle PR=\sqrt{36}=6\text{ cm}
\displaystyle \therefore PR=6\text{ cm}
\\

\displaystyle \textbf{Question 23: } \text{In the figure given below, medians }BD\text{ and }CE\text{ of }\triangle ABC
\displaystyle \text{meet at }G.\text{ Prove that:}
\displaystyle \text{(i) }\triangle EGD\sim\triangle CGB
\displaystyle \text{(ii) }BG=2GD\text{ from (i)}
\displaystyle \text{Answer:}
\displaystyle \text{Since }BD\text{ and }CE\text{ are medians,}
\displaystyle AD=DC\text{ and }AE=EB
\displaystyle \therefore E\text{ and }D\text{ are midpoints of }AB\text{ and }AC\text{ respectively.}
\displaystyle \therefore ED\parallel BC\quad\text{(midpoint theorem)}
\displaystyle \text{In }\triangle EGD\text{ and }\triangle CGB,
\displaystyle \angle DEG=\angle GCB\quad\text{(alternate angles, since }ED\parallel BC\text{)}
\displaystyle \angle EGD=\angle CGB\quad\text{(vertically opposite angles)}
\displaystyle \therefore \triangle EGD\sim\triangle CGB\quad\text{(AA similarity)}
\displaystyle \text{Hence, }\frac{GD}{GB}=\frac{ED}{BC}
\displaystyle \text{Now, in }\triangle ABC,\ E\text{ and }D\text{ are midpoints of }AB\text{ and }AC.
\displaystyle \therefore ED=\frac{1}{2}BC\quad\text{(midpoint theorem)}
\displaystyle \therefore \frac{ED}{BC}=\frac{1}{2}
\displaystyle \frac{GD}{GB}=\frac{1}{2}
\displaystyle \therefore GB=2GD
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.