Question

In figure, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that
$\frac{CA}{PA}=\frac{BC}{MP}$

Answer 1

In $△ABC$ and $△AMP$

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

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

using AA similarity

$△ABC\sim △AMP$

We know that if two triangles are similar, the ratio of their corresponding sides are proportional

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

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