CAT Similarity of Triangles - Practice Questions & MCQ

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EASY

In triangle PQR, S and T are points on sides PR and QR of $\triangle$ PQR such that $\angle P=\angle RTS$ then,

$\triangle R P Q \sim \triangle R T S$

$\triangle R P Q \cong \triangle R T S$

$\triangle R P Q \nsim \triangle R T S$

None of the above

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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:

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

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

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

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