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If angle 1 = angle 2 and ∆NSQ congrent to ∆MTR then prove that ∆PTS congrent to ∆PRQ

Posted by Mayur Patel 2 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Preeti Dabral 2 years, 2 months ago

Since $\triangle$ NSQ $\cong$ $\triangle$ MTR
$\therefore$ $\angle$ SQN = $\angle$ TRM
$\Rightarrow$ $\angle$ Q= $\angle$ R(in $\triangle$ PQR)
$\angle Q$ = 90^o- $\frac{1}{2}\angle P$
Again 1 = $\angle$ 2 [given in $\triangle$ PST](Isosceles property)
$\therefore$ $\angle$ 1= $\angle$ 2 = $\frac{1}{2}\left( {{{180}^ \circ } - \angle P} \right)$
$= {90^ \circ } - \frac{1}{2}\angle P$
Thus, in $\triangle$ PTS and $\triangle$ PRQ
$\angle$ 1 = $\angle$ Q [Each = 90° - $\frac{1}{2}\angle P$ ]
$\angle$ 2 = $\angle$ R, $\angle$ P= $\angle$ P(Common)
$\triangle$ PTS $\cong$ $\triangle$ PRQ

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