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If $Δ\ ABC$ and $Δ\ AMP$ are two right triangles, right angled at $B$ and $M$, respectively such that $∠MAP\ =\ ∠BAC$. Prove that:
(i) $ΔABC\ ∼\ ΔAMP$(ii) $\frac{CA}{PA}\ =\ \frac{BC}{MP}$
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Given:


$Δ\ ABC$ and $Δ\ AMP$ are two right triangles, right-angled at $B$ and $M$, respectively such that $∠MAP\ =\ ∠BAC$.


To do:


We have to prove that:


(i) $ΔABC\ ∼\ ΔAMP$


(ii) $\frac{CA}{PA}\ =\ \frac{BC}{MP}$


Solution:


(i) In $\vartriangle AMP$ and $\vartriangle ABC$, 

$\angle AMP = ∠ABC = 90^o$

$\angle MAP = \angle BAC$   (Vertically opposite angles)

Therefore,

$\vartriangle ABC ∼ \vartriangle AMP$     (By AA similarity)

(ii) $\vartriangle ABC ∼ \vartriangle AMP$

This implies,

$\frac{CA}{PA} = \frac{BC}{MP}$  (Corresponding sides are proportional)

Hence proved.

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Updated on: 10-Oct-2022

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