Question

In the adjoining figure $\angle ABC={90}^o$ and $\angle AMP={90}^o$
Prove that
(i) $△ABC\sim △AMP$
(ii) $\frac{CA}{PA}=\frac{BC}{MP}$

Answer 1

In the given figure, $△ABC$ and $△AMP$ are right angled triangles.

(i) $\angle ABC=\angle AMP={90}^0$ (Given)

$\angle CAB=\angle MAP$ (Common angles)

$\angle ACB=\angle APM$ (Remaining angles)

Therefore, $△ABC\sim △AMP$

(ii) When the triangles are equiangular then their corresponding sides are proportional, therefore,

$\frac{CA}{PA}=\frac{BC}{MP}$

Hence proved.