Top MBA Colleges in India Placement Wise

CAT Similarity of Triangles - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Concepts Covered - 1

Similarity of Triangles

AAA (Angle–Angle–Angle) similarity of two triangles.

If in two triangles, the corresponding angles are equal, then their corresponding sides are proportional and hence the triangles are similar.

Given: $\triangle A B C \text { and } \triangle D E F \text { such that } \angle A=\angle D, \angle B=\angle E \text { and } \angle C=\angle F$

To prove: $\triangle A B C \sim \triangle D E F$

Construction: Cut DP = AB and DQ = AC. Join PQ.

Proof:

$\text{In }\triangle A B C\text{ and } \triangle D P Q,\text{ we have}$

$A B=D P \qquad\text{[by construction]}$

$\angle A=\angle D \qquad \text { [given}]$

$A C=D Q \qquad\text{[by construction]}$

$\therefore \quad \triangle A B C \cong \triangle D P Q \quad \text { [by SAS-congruence }]$

$\Rightarrow \quad \angle B=\angle P$

$\Rightarrow \angle E=\angle P \qquad [\because \angle B=\angle E \text{(given)}]$

$\Rightarrow \quad P Q \| E F\qquad[\because \text { corresponding } \Delta \text { are equal}]$

$\Rightarrow \quad \frac{D P}{D E}=\frac{D Q}{D F}$

$\\ \Rightarrow \quad \frac{A B}{D E}=\frac{C A}{F D} \qquad[\because D P=A B \text { and } D Q=A C]$

$\text { Similarly, } \frac{A B}{D E}=\frac{B C}{E F} \\$

$\therefore \quad \frac{A B}{D E}=\frac{B C}{E F}=\frac{C A}{F D}$

$\text { Thus, } \angle A=\angle D, \;\angle B=\angle E,\; \angle C=\angle F \text { and } \frac{A B}{D E}=\frac{B C}{E F}=\frac{C A}{F D} \\$

$\text { Hence, } \triangle A B C \sim \triangle D E F$

Note: If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal.

SSS (Side–Side–Side) similarity for two triangles.

If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

Given: $\triangle A B C \text { and } \triangle D E F \text { in which } \frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}$

To Prove: $\triangle A B C \sim \triangle D E F$

Construction:

$\\\text{Let in } \triangle A B C\text{ and }\triangle D E F,\; \frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}(<1\;\;\text{and cut } D P=A B\text{ and } D Q=A C.\text{ Join }PQ.$

Proof: $\frac{A B}{D E}=\frac{A C}{D F} \Rightarrow \frac{D P}{D E}=\frac{D Q}{D F} \qquad[\because A B=D P \text { and } A C=D Q]$

So, by the converse of Thales' theorem, $PQ \| E F$

$\\\therefore \quad \angle P=\angle E \qquad \text { [ corresponding } \angle_S\; ]\\$

$\therefore\quad\angle Q=\angle F \qquad \text { [ corresponding } \angle_S\;]$

$\\\therefore \quad \triangle D P Q \sim \triangle D E F \qquad[\text { by AAA-similarity }]$

$\Rightarrow \quad \frac{D P}{D E}=\frac{P Q}{E F} \\$

$\Rightarrow \quad \frac{A B}{D E}=\frac{P Q}{E F}\qquad\ldots (i) \qquad[\because D P=A B]$

$\text {But, } \frac{A B}{D E}=\frac{B C}{E F}\qquad\ldots (ii) \qquad\text{[ given ]}$

$\Rightarrow \quad B C=P Q \\$

$\text { Thus, } A B=D P, A C=D Q \text { and } B C=P Q$

$\therefore \quad \triangle A B C \cong \triangle D P Q\qquad \text { [ by SSS-congruence ] } \\$

$\therefore \quad \angle A=\angle D, \angle B=\angle P=\angle E \text { and } \angle C=\angle Q=\angle F$

$\Rightarrow \quad \angle A=\angle D, \angle B=\angle E\text{ and }\angle C=\angle F$

Thus, the given triangles are equiangular and hence similar.

SAS (Side–Angle–Side) similarity for two triangles.

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Given: $\triangle A B C \text { and } \triangle D E F \text { in which } \angle A=\angle D \text { and } \frac{A B}{D E}=\frac{A C}{D F}$

To Prove: $\triangle A B C \sim \triangle D E F$

Construction: Cut DP = AB and DQ = AC. Join PQ.

Proof: $\text{In } \triangle A B C\text{ and }\triangle D P Q,\text{ we have}$

$A B=D P \qquad [\text{ by construction }]$

$\angle A=\angle D \qquad\text{ ( given )}$

$A C=D Q\qquad[\text{ by construction }]$

$\therefore \quad \triangle A B C \cong \triangle D P Q \qquad \text { [by SAS-congruence] }\\$

$\therefore \quad \angle A=\angle D,\; \angle B=\angle P \text { and } \angle C=\angle Q$

$\text { Now, } \frac{A B}{D E}=\frac{A C}{D F} \quad \text { (given) }\\$

$\Rightarrow \quad \frac{D P}{D E}=\frac{D Q}{D F}\qquad[\because A B=D P \text { and } A C=D Q]$

$\Rightarrow P Q \| E F \qquad \text{[ by the converse of Thales' theorem ]}$

$\Rightarrow \quad \angle P=\angle E\text{ and } \angle Q=\angle F \quad[\text { corresponding } \angle_S\;]$

$\text{Thus, } \angle A=\angle D,\; \angle B=\angle E\text{ and } \angle C=\angle F$

So, the given triangles are equiangular and hence similar.

Right Angle Theorem

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other.

Given: $\triangle A B C \text { in which } \angle B A C=90^{\circ} \text { and } A D \perp B C$

To Prove: $(i) \triangle D B A \sim \triangle A B C \quad(ii) \triangle D A C \sim \triangle A B C \quad(iii) \triangle D B A \sim \triangle D A C$

Proof:

$\\(i)\text{ In } \triangle D B A\text{ and }\triangle A B C,\text{ we have} \\\\ \angle B D A=\angle B A C=90^{\circ} \\ \\\angle D B A=\angle A B C \qquad \text { (common) } \\\\ \therefore \quad \triangle D B A \sim \triangle A B C \qquad\text{(by AA-similarity)}$

$\\(ii)\text{ In } \triangle D AC\text{ and }\triangle A B C,\text{ we have} \\\\ \angle C D A=\angle B A C=90^{\circ} \\ \\\angle D C A=\angle A C B \qquad \text { (common) } \\\\ \therefore \quad \triangle D AC \sim \triangle A B C \qquad\text{(by AA-similarity)}$

$\\(iii)\text{ In } \triangle D BA\text{ and }\triangle DAC,\text{ we have}$

${\color{Blue} \because \left.\begin{array}{l} \angle B+\angle B A D=90^{\circ} \\ \angle C+\angle C A D=90^{\circ} \quad \quad \quad \Rightarrow \angle B=\angle C A D \text { and } \angle C=\angle B A D \\ \angle B A D+\angle C A D=90^{\circ} \end{array}\right\}}$

$\\\\ \angle ADB=\angle C DA=90^{\circ}$

$\angle B=\angle C A D \\$

$\angle B A D=\angle C \\$

$\therefore \quad \triangle D B A \sim \triangle D A C \qquad\text { [by AAA-similarity] }$

Study other Related Concepts

CAT Similarity of Triangles Current Topic

CAT Basic Terms and Definitions of Lines and Angles

CAT Various Types of Angle

CAT Linear Pair of Angles

CAT Vertically Opposite Angles

CAT Corresponding Angle Axioms

CAT Alternate Interior Angle Theorem

CAT Interior Angle Theorem

CAT Parallel and a Transversal Lines

CAT Introduction to Triangles

CAT Congruence of Triangles

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

CAT Articles

CAT 2024 Latest News: Live Updates, Response Sheet, Answer Keys, Expected Cutoff, Results

Dec 17, 2024

CAT 2024 Login: CAT Answer Key @iimcat.ac.in Login Working Now

Dec 17, 2024

CAT Exam Dates 2024 (Announced) - Check Complete Schedule, Slot Timings

Dec 16, 2024

CAT 2024: Result Date (SOON), Final Answer Key, Cutoff, IIM Admission Process, Top Colleges

Dec 17, 2024

CAT 2024 Slot 3 Answer Key and Response Sheet (Released): Download and Check Solutions

Dec 17, 2024

CAT Score vs Percentile 2024, 2023, 2022: Know What CAT Percentile Will Your Score Turn Into

Dec 17, 2024

CAT Score vs Percentile 2024: CAT Marks vs Percentile Section-Wise, Expected & Previous Year 2023, 2022, 2021

Dec 17, 2024

CAT Percentile Predictor 2024 - How CAT 2024 Percentile Calculator Can Shape Your GD-PI Strategy

Dec 17, 2024

MBA/PGDM Admissions OPEN

Great Lakes PGPM & PGDM 2025

Apply

Admissions Open | Globally Recognized by AACSB (US) & AMBA (UK) | 17.3 LPA Avg. CTC for PGPM 2024 | Application Deadline: 1st Dec 2024

TAPMI Jaipur MBA Admissions 2025

Apply

67 Years of Established Legacy | Ranked #79 in India by NIRF | Among the only few top 100 B-Schools currently accepting applications

ISBR Business School PGDM Admissions 2025

Apply

180+ Companies | Highest CTC 15 LPA | Average CTC 8 LPA | Ranked as Platinum Institute by AICTE for 6 years in a row | Awarded Best Business School of the Year