Question

In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that
$\left(i\right)\Delta ABC\sim \Delta AMP$

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

Answer 1

(i) In$\Delta ABC$ and $\Delta AMP$, we have
$\angle A=\angle A$ (common angle)
$\angle ABC=\angle AMP={90}^{\circ }\left(Each{90}^{\circ }\right)$
$\therefore \Delta ABC\sim \Delta AMP$ (By AAA similarity criterion)

(ii) In$\Delta ABC$ and $\Delta AMP$, we have
$\angle A=\angle A$ (common angle)
$\angle ABC=\angle AMP={90}^{\circ }\left(Each{90}^{\circ }\right)$
$\therefore \Delta ABC\sim \Delta AMP$ (By AAA similarity criterion)
As, $\Delta ABC\sim AMP$ ,we know that,
If two triangles are similar then the corresponding sides are equal.
Hence,$\frac{CA}{PA}=\frac{BC}{MP}$