Question

In the given figure, $\Delta$ABC and $\Delta$AMP are right angled at B and M respectively.
Given AC$=10$cm, AP $=15$cm and PM$=12$ cm.
Prove that : $\Delta$ABC $\sim \Delta$ AMP.

Answer 1

In $△ABC,△AMP$

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

$\angle MAP=\angle BAC=\angle A$

As two angles are equal automatically third angle will be equal since sum of angles in a triangle is constant

By $AAA$ similarity

$△ABC\sim △AMP$