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Question 3In ...

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Question 3
In figure, if $∠ 1 = ∠ 2$ and $Δ N S Q ≅ Δ M T R$ , then prove that $Δ P T S \sim Δ P R Q$ .

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Solution

$G i v e n, Δ N S Q ≅ Δ M T R a n d ∠ 1 = ∠ 2$
To prove : $Δ P T S \sim Δ P R Q$
Proof :

$S i n c e, Δ N S Q ≅ Δ M T R$
So, SQ = TR ...(i)
$A l s o, ∠ 1 = ∠ 2 \Rightarrow P T = P S \dots (i i) [s i n c e, s i d e s o p p o s i t e t o e q u a l a n g l e s a r e a l s o e q u a l]$

Dividing Eqn. (ii) by (i), we get,
$\frac{P S}{S Q} = \frac{P T}{T R}$
$\Rightarrow S T ∥ Q R [b y c o n v e r s e o f b a s i c p r o p o r t i o n a l l y t h e o r e m]$
$∴ ∠ 1 = ∠ P Q R$
and $∠ 2 = ∠ P R Q$

$I n Δ P T S a n d Δ P R Q$ ,
$∠ P = ∠ P [c o m m o n a n g l e s]$
$∠ 1 = ∠ P Q R (C o r r e s p o n d i n g a n g l e s)$
$∠ 2 = ∠ P R Q (C o r r e s p o n d i n g a n g l e s)$
$∴ Δ P T S \sim Δ P R Q [b y A A A s i m i l a r i t y c r i t e r i o n]$

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