Question

In Fig., $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that:
$△ABC\sim △AMP$.

Answer 1

REF.Image.

Given : ABC and AMP are two right triangles,

$\angle ABC={90}^{\circ }$

$\angle AMP={90}^{\circ }$

Proof :-

In $△ABC$ and $△AMP$

$\angle CAB=\angle MAP$ (common angle)

$\angle ABC=\angle AMP$ (both ${90}^{\circ }$)

using AA similarity,

So, $△ABC\sim △AMP$