Class 10: Similarity of Triangles – CBSE Board Problems

$\displaystyle \textbf{Question 1. }\text{The corresponding sides of }\triangle ABC\text{ and }\triangle PQR\text{ are in the ratio }$
$\displaystyle 3:5. \ AD\perp BC\text{ and }PS\perp QR\text{ as shown in the following figures:} \hspace{0.2cm}\text{[CBSE 2025]}$ $\displaystyle \text{(i) Prove that }\triangle ADC\sim \triangle PSR$
$\displaystyle \text{(ii) If }AD=4\text{ cm, find the length of }PS.$
$\displaystyle \text{(iii) Using (ii) find }\mathrm{ar}(\triangle ABC):\mathrm{ar}(\triangle PQR)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: Corresponding sides of }\triangle ABC\text{ and }\triangle PQR\text{ are in ratio }3:5$
$\displaystyle \therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{3}{5},\ AD\perp BC\text{ and }PS\perp QR$
$\displaystyle \text{(i) }\triangle ADC\sim\triangle PSR\ \text{(To prove)}$
$\displaystyle \text{In }\triangle ADC\text{ and }\triangle PSR$
$\displaystyle \angle C=\angle R\qquad [\because \triangle ABC\sim\triangle PSR]$
$\displaystyle \angle D=\angle S=90^\circ$
$\displaystyle \therefore \triangle ADC\sim\triangle PSR$
$\displaystyle \text{(ii)}\ \frac{AD}{PS}=\frac{AB}{PQ}$
$\displaystyle \Rightarrow \frac{AD}{PS}=\frac{3}{5}$
$\displaystyle \Rightarrow \frac{4}{PS}=\frac{3}{5}$
$\displaystyle \Rightarrow PS=\frac{20}{3}\ \text{cm}$
$\displaystyle \text{(iii)}\ \frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)}=\frac{\frac{1}{2}AD\times BC}{\frac{1}{2}PS\times QR}$
$\displaystyle \Rightarrow \frac{1}{2}\times\frac{4}{\frac{20}{3}}\times\left(\frac{3}{5}\right)^2$
$\displaystyle =\frac{3^2}{5^2}=\frac{9}{25}$
$\\$

$\displaystyle \textbf{Question 2. }\text{A }1.5\text{ m tall boy is walking away from the base of a lamp post which is}$
$\displaystyle 12\text{ m high, at the speed of }2.5\text{ m/sec. Find the length of his shadow after }3\text{ seconds.}$
$\displaystyle \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let after 3 seconds boy at }D.$
$\displaystyle \text{Length of shadow }=DE=x\ \text{m}$
$\displaystyle \text{In }\triangle ABE\text{ and }\triangle CDE$
$\displaystyle \angle E=\angle E\ (\text{Common})$
$\displaystyle \angle ABE=\angle CDE=90^\circ$
$\displaystyle \therefore \triangle ABE\sim\triangle CDE$
$\displaystyle \therefore \frac{AB}{CD}=\frac{BE}{DE}$
$\displaystyle \therefore \frac{12}{1.5}=\frac{7.5+x}{x}$
$\displaystyle \therefore 12x=1.5(7.5+x)$
$\displaystyle \therefore \frac{120}{15}x=7.5+x$
$\displaystyle \therefore 8x-x=7.5$
$\displaystyle \therefore 7x=7.5$
$\displaystyle \therefore x=\frac{75}{70}=\frac{15}{14}$
$\displaystyle \therefore x=1.071\ \text{m}$
$\displaystyle \therefore \text{Length of his shadow after 3 sec is }1.071\text{ m}$
$\\$

$\displaystyle \textbf{Question 3. }\text{In parallelogram }ABCD,\text{ side }AD\text{ is produced to a point }E\text{ and }BE$
$\displaystyle \text{intersects }CD\text{ at }F. \text{Prove that }\triangle ABE\sim \triangle CFB. \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To prove: }\triangle ABE\sim\triangle CFB$
$\displaystyle \text{Proof: As given that }ABCD\text{ is a parallelogram}$
$\displaystyle \therefore AE\parallel BC\text{ and }AB\parallel DC$
$\displaystyle \text{In }\triangle ABE\text{ and }\triangle CFB$
$\displaystyle \angle ABE=\angle FBC\ [\text{A.I.A.}]$
$\displaystyle \angle AEB=\angle FCB\ [\text{A.I.A.}]$
$\displaystyle \therefore \triangle ABE\sim\triangle CFB$
$\displaystyle \text{So, by AA criteria}$
$\displaystyle \text{Hence proved}$
$\\$

$\displaystyle \textbf{Question 4. }\text{State basic proportionality theorem. Use it to prove the following:}$
$\displaystyle \text{If three parallel lines }l,m,n\text{ are intersected by transversals }q\text{ and }s\text{ as shown}$
$\displaystyle \text{in the adjoining figure, then }\frac{AB}{BC}=\frac{DE}{EF}. \hspace{0.2cm}\text{[CBSE 2025]}$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Basic Proportionality Theorem: If a line is drawn parallel}$
$\displaystyle \text{to one side of a triangle and intersects the other two}$
$\displaystyle \text{sides, then it divides those two sides in the same ratio.}$
$\displaystyle \text{Here }l\parallel m\parallel n\text{ and }q\text{ \& }s\text{ are transversal lines.}$
$\displaystyle \text{Now in }C\text{ to }D$
$\displaystyle \text{In }\triangle ACD,\ BG\parallel AD$
$\displaystyle \therefore \frac{CB}{AB}=\frac{CG}{GD}\qquad (i)$
$\displaystyle \text{In }\triangle CDF$
$\displaystyle GE\parallel CF$
$\displaystyle \therefore \frac{DG}{GC}=\frac{DE}{EF}$
$\displaystyle \therefore \frac{CG}{DG}=\frac{EF}{DE}\qquad (ii)$
$\displaystyle \text{From }(i)\text{ and }(ii),$
$\displaystyle \frac{CB}{AB}=\frac{EF}{DE}$
$\displaystyle \therefore \frac{AB}{BC}=\frac{DE}{EF}$
$\\$

$\displaystyle \textbf{Question 5. }\triangle ABC\text{ and }\triangle PQR\text{ are shown in the adjoining figures.}$
$\displaystyle \text{The measure of }\angle C\text{ is:} \hspace{0.2cm}\text{[CBSE 2025]}$ $\displaystyle \text{(a) }140^{\circ} \qquad \text{(b) }80^{\circ}$
$\displaystyle \text{(c) }60^{\circ} \qquad \text{(d) }40^{\circ}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(d)}\ \text{Here }\frac{RP}{AC}=\frac{RQ}{AB}=\frac{PQ}{BC}$
$\displaystyle \therefore \triangle ABC\sim\triangle RQP$
$\displaystyle \therefore \angle C=\angle P=40^\circ$
$\\$

$\displaystyle \textbf{Question 6. }E\text{ and }F\text{ are points on the sides }AB\text{ and }AC\text{ respectively } \text{of a }\triangle ABC$
$\displaystyle \text{ such that }\frac{AE}{EB}=\frac{AF}{FC}=\frac{1}{2}. \text{Which of the following } \text{relation is true?} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{(a) }EF=2BC \qquad \text{(b) }BC=2EF$
$\displaystyle \text{(c) }EF=3BC \qquad \text{(d) }BC=3EF$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(d)}$
$\displaystyle \triangle AEF\sim\triangle ABC$
$\displaystyle \therefore \frac{AE}{AB}=\frac{AF}{AC}=\frac{EF}{BC}$
$\displaystyle \therefore \frac{1}{3}=\frac{EF}{BC}$
$\displaystyle \therefore BC=3EF$
$\\$

$\displaystyle \textbf{Question 7. }\text{If a line is drawn parallel to one side of a triangle to intersect the other}$
$\displaystyle \text{two sides in distinct points, then prove that the other two sides are divided in the}$
$\displaystyle \text{same ratio.} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: A triangle }ABC,\text{ in which }DE\parallel BC,\text{ meeting}$
$\displaystyle AB\text{ at }D\text{ and }AC\text{ at }E$
$\displaystyle \text{To Prove: }\frac{AD}{DB}=\frac{AE}{EC}$
$\displaystyle \text{Construction: Join }BE,\ CD\text{ and draw }EL\perp AD$
$\displaystyle \text{Proof: }\triangle BDE\text{ and }\triangle CDE\text{ are on the same base }DE$
$\displaystyle \text{and between the same parallel lines }BC\text{ and }DE,\text{ hence equal}$
$\displaystyle \text{in area, i.e.,}$
$\displaystyle \text{ar}(\triangle BDE)=\text{ar}(\triangle CDE)\qquad (i)$
$\displaystyle \text{Now,}$
$\displaystyle \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)}=\frac{\frac{1}{2}AD\cdot EL}{\frac{1}{2}BD\cdot EL}=\frac{AD}{BD}\qquad (ii)$
$\displaystyle \text{Similarly,}$
$\displaystyle \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)}=\frac{\frac{1}{2}AE\cdot DP}{\frac{1}{2}EC\cdot DP}=\frac{AE}{EC}\qquad (iii)$
$\displaystyle \text{Also,}$
$\displaystyle \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)}=\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)}\qquad [\text{Using }(i)]$
$\displaystyle \therefore \frac{AD}{BD}=\frac{AE}{EC}\qquad [\text{From }(ii)\text{ and }(iii)]$
$\displaystyle \therefore \frac{AD}{DB}=\frac{AE}{EC}$
$\\$

$\displaystyle \textbf{Question 8. }\text{In the given figure, }ABCD\text{ is a quadrilateral. Diagonal }BD\text{ bisects}$
$\displaystyle \angle B\text{ and }\angle D\text{ both. Prove that:}$
$\displaystyle \text{(i) }\triangle ABD\sim \triangle CBD$
$\displaystyle \text{(ii) }AB=BC \hspace{0.2cm}\text{[CBSE 2024]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Since, }BD\text{ bisects }\angle B\text{ and }\angle D\text{ then}$
$\displaystyle \angle ABD=\angle CBD$
$\displaystyle \angle ADB=\angle BDC$
$\displaystyle \text{(i) In }\triangle ABD\text{ and }\triangle CBD,$
$\displaystyle \angle ABD=\angle CBD$
$\displaystyle \angle ADB=\angle BDC$
$\displaystyle \therefore \triangle ABD\sim\triangle CBD\ (\text{AA similarity})$
$\displaystyle \text{(ii) Now, }\triangle ABD\sim\triangle CBD,$
$\displaystyle \therefore \frac{AB}{BC}=\frac{BD}{BD}=\frac{AD}{CD}$
$\displaystyle \text{(Corresponding sides of similar triangles are proportional)}$
$\displaystyle \therefore \frac{AB}{BC}=\frac{BD}{BD}$
$\displaystyle \therefore \frac{AB}{BC}=1$
$\displaystyle \therefore AB=BC$
$\\$

$\displaystyle \textbf{Question 9. }\text{In the given figure }PA,QB\text{ and }RC\text{ are each perpendicular to }AC.$
$\displaystyle \text{If }AP=x,\ BQ=y\text{ and }CR=z,\text{ then prove that }\frac{1}{x}+\frac{1}{z}=\frac{1}{y}.$
$\displaystyle \hspace{0.2cm}\text{[CBSE 2024]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle PAC\text{ and }\triangle QBC$
$\displaystyle \angle PAC=\angle QBC\qquad [90^\circ\ \text{each}]$
$\displaystyle \angle PCA=\angle QCB\qquad [\text{Common}]$
$\displaystyle \therefore \triangle PAC\sim\triangle QBC\qquad [\text{AA similarity}]$
$\displaystyle \Rightarrow \frac{PA}{QB}=\frac{AC}{BC}=\frac{PC}{QC}$
$\displaystyle \Rightarrow \frac{PA}{QB}=\frac{AC}{BC}$
$\displaystyle \Rightarrow \frac{x}{y}=\frac{AC}{BC}\ \text{or}\ \frac{y}{x}=\frac{BC}{AC}\qquad (i)$
$\displaystyle \text{[Corresponding sides of similar triangles are proportional]}$
$\displaystyle \text{In }\triangle RCA\text{ and }\triangle QBA$
$\displaystyle \angle RCA=\angle QBA\qquad [90^\circ\ \text{Each}]$
$\displaystyle \angle RAC=\angle QAB\qquad [\text{Common}]$
$\displaystyle \therefore \triangle RCA\sim\triangle QBA\qquad [\text{AA similarity}]$
$\displaystyle \Rightarrow \frac{RC}{QB}=\frac{AC}{AB}=\frac{RA}{QA}$
$\displaystyle \Rightarrow \frac{RC}{QB}=\frac{AC}{AB}$
$\displaystyle \Rightarrow \frac{z}{y}=\frac{AC}{AB}\ \text{or}\ \frac{y}{z}=\frac{AB}{AC}\qquad (ii)$
$\displaystyle \text{Adding (i) and (ii), we get}$
$\displaystyle \frac{y}{x}+\frac{y}{z}=\frac{BC}{AC}+\frac{AB}{AC}$
$\displaystyle \Rightarrow \frac{y}{x}+\frac{y}{z}=\frac{BC+AB}{AC}=\frac{AC}{AC}=1$
$\displaystyle \Rightarrow y\left(\frac{1}{x}+\frac{1}{z}\right)=1$
$\displaystyle \Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{1}{y}$
$\displaystyle \therefore \text{Hence proved}$
$\\$

$\displaystyle \textbf{Question 10. }\text{In the given figure, }\triangle ABC\sim \triangle QPR. \text{If }AC=6\text{ cm, }$
$\displaystyle BC=5\text{ cm,} \ QR=3\text{ cm and }PR=x;\text{ then the value of }x\text{ is:} \hspace{0.2cm}\text{[CBSE 2023]}$ $\displaystyle \text{(a) }3.6\text{ cm} \qquad \text{(b) }2.5\text{ cm}$ $\displaystyle \text{(c) }10\text{ cm} \qquad \text{(d) }3.2\text{ cm}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(b)}$
$\displaystyle \triangle ABC\sim\triangle QPR$
$\displaystyle \therefore \frac{AC}{QR}=\frac{BC}{PR}$
$\displaystyle \therefore \frac{6}{3}=\frac{5}{x}$
$\displaystyle \therefore x=2.5\ \text{cm}$
$\\$

$\displaystyle \textbf{Question 11. }\text{In }\triangle ABC,\ PQ\parallel BC. \text{If }PB=6\text{ cm, }AP=4\text{ cm, }$
$\displaystyle AQ=8\text{ cm,} \ \text{find the length of }AC. \hspace{0.2cm}\text{[CBSE 2023]}$ $\displaystyle \text{(a) }12\text{ cm} \qquad \text{(b) }20\text{ cm}$
$\displaystyle \text{(c) }6\text{ cm} \qquad \text{(d) }14\text{ cm}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(b)}$
$\displaystyle \text{In }\triangle ABC,\ PQ\parallel BC$
$\displaystyle \therefore \frac{AP}{BP}=\frac{AQ}{QC}\qquad[\text{BPT}]$
$\displaystyle \therefore \frac{4}{6}=\frac{8}{QC}$
$\displaystyle \therefore QC=12\ \text{cm}$
$\displaystyle \therefore AC=AQ+QC=8+12=20\ \text{cm}$
$\\$

$\displaystyle \textbf{Question 12. }\text{In the given figure, }DE\parallel BC\text{ and all measurements are given}$
$\displaystyle \text{in centimetres. The length of }AE\text{ is:} \hspace{0.2cm}\text{[CBSE 2023(C)]}$ $\displaystyle \text{(a) }2\text{ cm} \qquad \text{(b) }2.25\text{ cm}$
$\displaystyle \text{(c) }2.5\text{ cm} \qquad \text{(d) }2.75\text{ cm}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(b)}$
$\displaystyle \text{In }\triangle ABC,\ \text{we have }DE\parallel BC$
$\displaystyle \therefore \frac{AD}{DB}=\frac{AE}{EC}\qquad[\text{BPT}]$
$\displaystyle \therefore \frac{3}{4}=\frac{AE}{3}$
$\displaystyle \therefore AE=\frac{9}{4}=2.25\ \text{cm}$
$\\$

$\displaystyle \textbf{Question 13. }\text{In the given figure, }AB\perp BC\text{ and }DE\perp AC. \text{Prove that}$
$\displaystyle \triangle ABC\sim \triangle AED. \hspace{0.2cm}\text{[CBSE 2023(C)]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ABC\text{ and }\triangle AED,$
$\displaystyle \angle A=\angle A\qquad (\text{common})$
$\displaystyle \angle ABC=\angle AED\qquad (90^\circ\ \text{each})$
$\displaystyle \therefore \triangle ABC\sim\triangle AED\qquad (\text{AA similarity})$
$\\$

$\displaystyle \textbf{Question 14. }D\text{ is a point on the side }BC\text{ of a triangle }ABC\text{ such that}$
$\displaystyle \angle ADC=\angle BAC,\text{ prove that }CA^{2}=CB\cdot CD. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ADC\text{ and }\triangle BAC$
$\displaystyle \angle ADC=\angle BAC\qquad (\text{given})$
$\displaystyle \angle C=\angle C\qquad (\text{Common})$
$\displaystyle \therefore \triangle ADC\sim\triangle BAC\qquad (\text{AA similarity})$
$\displaystyle \Rightarrow \frac{AD}{BA}=\frac{DC}{AC}=\frac{AC}{BC}$
$\displaystyle \text{(Corresponding sides of similar triangles are proportional)}$
$\displaystyle \Rightarrow \frac{DC}{AC}=\frac{AC}{BC}$
$\displaystyle \Rightarrow AC^2=CB\times CD$
$\\$

$\displaystyle \textbf{Question 15. }\text{If }AD\text{ and }PM\text{ are medians of triangles }ABC\text{ and }PQR$
$\displaystyle \text{respectively where }\triangle ABC\sim \triangle PQR,\text{ prove that }\frac{AB}{PQ}=\frac{AD}{PM}. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \triangle ABC\sim\triangle PQR\qquad (\text{given})$
$\displaystyle \angle B=\angle Q\qquad [\text{Corresponding angles of similar triangles are equal}] \qquad (i)$
$\displaystyle \text{and}$
$\displaystyle \frac{AB}{PQ}=\frac{BC}{QR}\qquad [\text{Corresponding sides of similar triangles are proportional}]$
$\displaystyle \text{and since }D\text{ is midpoint of }BC\text{ and }M\text{ is midpoint of }QR$
$\displaystyle \Rightarrow \frac{AB}{PQ}=\frac{2BD}{2QM}$
$\displaystyle \Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}\qquad (ii)$
$\displaystyle \text{In }\triangle ABD\text{ and }\triangle PQM$
$\displaystyle \frac{AB}{PQ}=\frac{BD}{QM}\qquad [\text{from }(ii)]$
$\displaystyle \angle B=\angle Q\qquad [\text{from }(i)]$
$\displaystyle \therefore \triangle ABD\sim\triangle PQM\qquad (\text{SAS})$
$\displaystyle \Rightarrow \frac{AB}{PQ}=\frac{AD}{PM}$
$\\$

$\displaystyle \textbf{Question 16. }\text{In }\triangle ABC\text{ and }\triangle DEF,\ \angle F=\angle C,\ \angle B=\angle E\text{ and}$
$\displaystyle AB=\frac{1}{2}DE. \text{Then, the two triangles are} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) Congruent, but not similar.} \qquad \text{(b) Similar, but not congruent.}$
$\displaystyle \text{(c) Neither congruent nor similar.} \qquad \text{(d) Congruent as well as similar.}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(b)}$
$\displaystyle \text{In }\triangle ABC\text{ and }\triangle DEF$
$\displaystyle \angle B=\angle E\ (\text{Given})$
$\displaystyle \angle C=\angle F\ (\text{Given})$
$\displaystyle \therefore \triangle ABC\sim\triangle DEF\ (\text{AA similarity})$
$\displaystyle \text{But }\triangle ABC\text{ is not congruent to }\triangle DEF\text{ as }AB\neq DE$
$\\$

$\displaystyle \textbf{Question 17. }\text{In fig., }PA,QB\text{ and }RC\text{ are each perpendicular to }AC. \text{ If }$
$\displaystyle x=8\text{ cm} \ \text{and }z=6\text{ cm, then }y\text{ is equal to} \hspace{0.2cm}\text{[CBSE 2021]}$
$\displaystyle \text{(a) }\frac{56}{7}\text{ cm} \qquad \text{(b) }\frac{7}{56}\text{ cm}$
$\displaystyle \text{(c) }\frac{25}{7}\text{ cm} \qquad \text{(d) }\frac{24}{7}\text{ cm}$ $\displaystyle \text{Answer:}$
$\displaystyle \textbf{(d)}$
$\displaystyle \text{In }\triangle PAC\text{ and }\triangle QBC,$
$\displaystyle \angle PAC=\angle QBC\qquad (90^\circ\ \text{each})$
$\displaystyle \angle PCA=\angle QCB\qquad (\text{common})$
$\displaystyle \therefore \triangle PAC\sim\triangle QBC\qquad (\text{AA})$
$\displaystyle \therefore \frac{PA}{QB}=\frac{AC}{BC}=\frac{x}{y}=\frac{AC}{BC}$
$\displaystyle \therefore \frac{y}{x}=\frac{BC}{AC}\qquad (i)$
$\displaystyle \text{Similarly, }\triangle BAC\sim\triangle QAB\qquad (\text{AA})$
$\displaystyle \therefore \frac{y}{z}=\frac{AB}{AC}\qquad (ii)$
$\displaystyle \text{Adding }(i)\text{ and }(ii),\text{ we get}$
$\displaystyle y\left(\frac{1}{x}+\frac{1}{z}\right)=\frac{AB+BC}{AC}=\frac{AC}{AC}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{z}=\frac{1}{y}$
$\displaystyle \therefore \frac{1}{8}+\frac{1}{6}=\frac{1}{y}$
$\displaystyle \therefore y=\frac{24}{7}\ \text{cm}$
$\\$

$\displaystyle \textbf{Question 18. }\text{A farmer has a field in the shape of trapezium, whose map with scale}$
$\displaystyle 1\text{ cm}=20\text{ m, is given below. The field is divided into four parts by joining the}$
$\displaystyle \text{opposite vertices.} \hspace{0.2cm}\text{[CBSE 2021]}$ $\displaystyle \text{Based on the above information, answer the following questions:}$
$\displaystyle \text{(i) The two triangular regions }AOB\text{ and }COD\text{ are}$
$\displaystyle \text{(a) Similar by AA criterion}$
$\displaystyle \text{(b) Similar by SAS criterion}$
$\displaystyle \text{(c) Similar by RHS criterion}$
$\displaystyle \text{(d) Not similar}$
$\displaystyle \text{(ii) If the ratio of the perimeter of }\triangle AOB\text{ to the perimeter of }\triangle COD$
$\displaystyle \text{would have been }1:4,\text{ then}$
$\displaystyle \text{(a) }AB=2CD \qquad \text{(b) }AB=4CD$
$\displaystyle \text{(c) }CD=2AB \qquad \text{(d) }CD=4AB$
$\displaystyle \text{(iii) If in }\triangle AOD\text{ and }\triangle BOC,\ \frac{AO}{BC}=\frac{AD}{BO}=\frac{OD}{OC},$
$\displaystyle \text{then}$
$\displaystyle \text{(a) }\triangle AOD\sim \triangle BOC$
$\displaystyle \text{(b) }\triangle ADO\sim \triangle BCO$
$\displaystyle \text{(c) }\triangle AADO\sim \triangle BCO$
$\displaystyle \text{(d) }\triangle ODA\sim \triangle OBC$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i)(a)}$
$\displaystyle \text{In }\triangle AOB\text{ and }\triangle COD,$
$\displaystyle \angle ABO=\angle ODC\qquad [\text{Alternate interior angles, as }AB\parallel DC]$
$\displaystyle \angle AOB=\angle COD\qquad [\text{Vertically opposite angles}]$
$\displaystyle \therefore \triangle AOB\sim\triangle COD\qquad (\text{AA similarity})$

$\displaystyle \textbf{(ii)(d)}$
$\displaystyle \text{As, }\triangle AOB\sim\triangle COD,\text{ then}$
$\displaystyle \frac{\text{Perimeter of }\triangle AOB}{\text{Perimeter of }\triangle COD}=\frac{AB}{CD}=\frac{AO}{CO}=\frac{OB}{OD}$
$\displaystyle \Rightarrow \frac{1}{4}=\frac{AB}{CD}$
$\displaystyle \Rightarrow 4AB=CD$

$\displaystyle \textbf{(iii)(b)}$
$\displaystyle \text{We have,}$
$\displaystyle \frac{AO}{BC}=\frac{AD}{BO}=\frac{OD}{OC}$
$\displaystyle \therefore \triangle AOD\sim\triangle BCO\qquad (\text{SSS similarity})$
$\\$

$\displaystyle \textbf{Question 19. }\text{In the given figure, }DE\parallel AC\text{ and }DC\parallel AP. \text{Prove that}$
$\displaystyle \frac{BE}{EC}=\frac{BC}{CP}. \hspace{0.2cm}\text{[CBSE 2020]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ABC,\ DE\parallel AC$
$\displaystyle \therefore \frac{BD}{AD}=\frac{BE}{EC}\qquad (i)\ [\text{BPT}]$
$\displaystyle \text{In }\triangle ABP,\ CD\parallel AP$
$\displaystyle \therefore \frac{BD}{AD}=\frac{BC}{CP}\qquad (ii)\ [\text{BPT}]$
$\displaystyle \text{From (i) and (ii)}$
$\displaystyle \therefore \frac{BE}{EC}=\frac{BC}{CP}$
$\\$

$\displaystyle \textbf{Question 20. }\text{In the given figure, }\angle D=\angle E\text{ and }\frac{AD}{DB}=\frac{AE}{EC},\text{ prove}$
$\displaystyle \text{that }BAC\text{ is an isosceles triangle.} \hspace{0.2cm}\text{[CBSE 2020]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Given: In the given figure,}$
$\displaystyle \angle D=\angle E\ \text{and}\ \frac{AD}{DB}=\frac{AE}{EC}$
$\displaystyle \text{To prove: }\triangle BAC\text{ is an isosceles triangle}$
$\displaystyle \text{Proof: In }\triangle ADE,\ \angle D=\angle E\ (\text{given})$
$\displaystyle \therefore AE=AD\qquad (i)$
$\displaystyle \text{(sides opposite to equal angles)}$
$\displaystyle \text{Also, }\frac{AD}{DB}=\frac{AE}{EC}\ (\text{given})$
$\displaystyle \therefore DB=EC\qquad (ii)\ [\text{from }(i)]$
$\displaystyle \text{From (i) and (ii)}$
$\displaystyle AD+DB=AE+EC$
$\displaystyle \therefore AB=AC$
$\displaystyle \therefore \triangle BAC\text{ is an isosceles triangle}$
$\\$

$\displaystyle \textbf{Question 21. }\text{In Fig., }\angle ACB=90^{\circ}\text{ and }CD\perp AB,\text{ prove that}$
$\displaystyle CD^{2}=BD\times AD. \hspace{0.2cm}\text{[CBSE 2019]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle ADC\text{ and }\triangle ACB,$
$\displaystyle \angle A=\angle A\qquad (\text{common})$
$\displaystyle \angle ADC=\angle ACB=90^\circ\qquad (\text{given})$
$\displaystyle \therefore \triangle ADC\sim\triangle ACB\qquad (\text{by AA})$
$\displaystyle \therefore \frac{AD}{AC}=\frac{AC}{BC}\Rightarrow \frac{AD}{CD}=\frac{AC}{BC}\qquad (i)$
$\displaystyle \text{In }\triangle BCA\text{ and }\triangle BDC,$
$\displaystyle \angle CBA=\angle CBD\qquad (\text{common})$
$\displaystyle \angle BCA=\angle BDC=90^\circ$
$\displaystyle \therefore \triangle BCA\sim\triangle BDC\qquad (\text{by AA})$
$\displaystyle \therefore \frac{AC}{BD}=\frac{BC}{BC}=\frac{CD}{BD}\qquad (ii)$
$\displaystyle \text{Equating (i) and (ii)}$
$\displaystyle \therefore \frac{AD}{CD}=\frac{CD}{BD}$
$\displaystyle \therefore CD^2=BD\times AD$
$\\$

$\displaystyle \textbf{Question 22. }X\text{ is a point on the side }BC\text{ of }\triangle ABC.\ XM\text{ and }XN\text{ are drawn}$
$\displaystyle \text{parallel to }AB\text{ and }AC\text{ respectively meeting }AB\text{ in }N\text{ and }AC\text{ in }M.$
$\displaystyle MN\text{ produced meets }CB\text{ produced at }T. \text{Prove that }TX^{2}=TB\times TC.$
$\displaystyle \hspace{0.2cm}\text{[CBSE 2018(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }XM\parallel AB\text{ and }XN\parallel AC$
$\displaystyle X\text{ is any point on }BC$
$\displaystyle MN\text{ produced meeting }CB\text{ produced at }T$
$\displaystyle \text{To prove: }TX^2=TB\times TC$
$\displaystyle \text{Proof: In }\triangle TMC,$
$\displaystyle \frac{TN}{TM}=\frac{TX}{TC}\qquad (i)$
$\displaystyle \text{(BPT, as }NX\parallel MC)$
$\displaystyle \text{In }\triangle TMX,$
$\displaystyle \frac{TN}{TM}=\frac{TB}{TX}\qquad (ii)$
$\displaystyle \text{(BPT, as }BN\parallel MX)$
$\displaystyle \text{Equating (i) and (ii), we get}$
$\displaystyle \frac{TX}{TC}=\frac{TB}{TX}$
$\displaystyle \therefore TX^2=TB\times TC$
$\\$

$\displaystyle \textbf{Question 23. }\text{If }\triangle ABC\sim \triangle RPQ,\ AB=3\text{ cm, }BC=5\text{ cm, }AC=6\text{ cm,}$
$\displaystyle RP=6\text{ cm and }PQ=10\text{ cm, then find }QR. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \triangle ABC\sim\triangle RPQ\ (\text{Given})$
$\displaystyle \therefore \frac{AB}{RP}=\frac{BC}{PQ}=\frac{AC}{RQ}$
$\displaystyle \text{(Proportional sides of similar triangles)}$
$\displaystyle \therefore \frac{3}{6}=\frac{5}{10}=\frac{6}{QR}$
$\displaystyle \therefore \frac{1}{2}=\frac{6}{QR}$
$\displaystyle \therefore QR=12\ \text{cm}$
$\\$

$\displaystyle \textbf{Question 24. }R\text{ and }S\text{ are points on the sides }DE\text{ and }EF\text{ respectively of a}$
$\displaystyle \triangle DEF\text{ such that }ER=5\text{ cm, }RD=2.5\text{ cm, }SE=1.5\text{ cm and }$
$\displaystyle FS=3.5\text{ cm.} \ \text{Find whether }RS\parallel DF\text{ or not.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have }RE=5\ \text{cm and }RD=2.5\ \text{cm}$
$\displaystyle \therefore \frac{RE}{RD}=\frac{5}{2.5}=\frac{2}{1}$
$\displaystyle \text{Similarly, we have }ES=1.5\ \text{cm and }SF=3.5\ \text{cm}$
$\displaystyle \therefore \frac{ES}{SF}=\frac{1.5}{3.5}=\frac{3}{7}$
$\displaystyle \text{Now,}$
$\displaystyle \frac{RE}{RD}\neq\frac{ES}{SF}$
$\displaystyle \therefore RS\text{ is not parallel to }DF$
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$\displaystyle \textbf{Question 25. }\text{In the figure, }ABCD\text{ is a parallelogram and }E\text{ divides }BC\text{ in the ratio }$
$\displaystyle 1:3. \ DB\text{ and }AE\text{ intersect at }F. \text{Show that }DF=4FB\text{ and }AF=4FE. \hspace{0.2cm}\text{[CBSE 2016]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Given: }ABCD\text{ is a parallelogram and }BE:EC=1:3$
$\displaystyle \text{To show: }DF=4FB\text{ and }AF=4FE$
$\displaystyle \text{Proof: In }\triangle ADF\text{ and }\triangle EBF$
$\displaystyle \angle ADF=\angle EBF\qquad (\text{Alternate angles})$
$\displaystyle \angle AFD=\angle EFB\qquad (\text{V.O.A.})$
$\displaystyle \therefore \triangle ADF\sim\triangle EBF\qquad (\text{by AA})$
$\displaystyle \therefore \frac{DF}{BF}=\frac{AF}{EF}=\frac{AD}{BE}\qquad (i)$
$\displaystyle \text{As }\frac{BE}{EC}=\frac{1}{3}\Rightarrow EC=3BE$
$\displaystyle \therefore BC=BE+EC=BE+3BE$
$\displaystyle \therefore BC=4BE$
$\displaystyle \text{As }AD=BC\qquad [\text{Opposite sides of parallelogram are equal}]$
$\displaystyle \therefore AD=4BE$
$\displaystyle \text{Put in (i), we get}$
$\displaystyle \frac{DF}{BF}=\frac{AF}{EF}=\frac{4}{1}$
$\displaystyle \therefore DF=4BF\text{ and }AF=4EF$
$\\$

$\displaystyle \textbf{Question 26. }\text{In the figure, }PQR\text{ and }QST\text{ are two right triangles, right-angled}$
$\displaystyle \text{at }R\text{ and }T\text{ respectively. Prove that }QR\times QS=QP\times QT. \hspace{0.2cm}\text{[CBSE 2016]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Given: In }\triangle PQR,\ \angle R=90^\circ\text{ and in }\triangle QST,\ \angle T=90^\circ$
$\displaystyle \text{To prove: }QR\times QS=QP\times QT$
$\displaystyle \text{Proof: In }\triangle PQR\text{ and }\triangle SQT$
$\displaystyle \angle PQR=\angle SQT\qquad (\text{Common})$
$\displaystyle \angle PRQ=\angle STQ=90^\circ\qquad (\text{Given})$
$\displaystyle \therefore \triangle PQR\sim\triangle SQT\qquad (\text{By AA})$
$\displaystyle \Rightarrow \frac{QR}{QT}=\frac{QP}{QS}$
$\displaystyle \text{(Corresponding sides of similar triangles are proportional)}$
$\displaystyle \Rightarrow QR\times QS=QP\times QT$
$\\$