Question

$△ABC$ 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

Given figure : (Image )

i) To prove

$△ABC$ is ismilar to triangle AMP

We can prove this using angle-angle similarity postulate

In triangles AMP and ABC

$\angle AMP=\angle ABC=90°$

$\angle MAP=\angle BAC=$same angle (common angle)

Hence, two angles of the required triangles are same

$\therefore$By angle-angle (AA) similarity:

We can say that $△ABC\sim △AMP$

ii) To prove $\to \frac{CA}{PA}=\frac{BC}{MP}$

$\because$ In part i) we already proved

that $\angle AMP=\angle ABC=90°$

and $\angle MAP=\angle BAC=$ common angle

and sum of $3$ angles of a triangle$=180°$

$\therefore \angle APM=\angle ACB$

From side-angle-side similarity theorem,

we can that the sides including the angles which are equal must be proportional to each other

Sides including $\angle APM=AP$ and $MP$

Sides including $\angle ACB=AC$ & $BC$

Using S-A-S similarity theorem

$\frac{AP}{MP}=\frac{AC}{BC}$

$\Rightarrow \frac{BC}{MP}=\frac{AC}{AP}$

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

Hence proved.