Question

In Fig., $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that:
(i) $△ABC\sim △AMP$
(ii) $\frac{CA}{PA}=\frac{BC}{MP}$

Answer 1

In $△ABC$ and $△AMP,$

$\angle ABC=\angle AMP={90}^o$

$\angle A$ is common

$\therefore$ By AA criterion of similarity, $△ABC$ $\sim$ $△AMP$

$\therefore$ $\frac{CA}{PA}=\frac{BC}{MP}$ (Corresponding Sides of Similar Triangles)

Hence Proved