# In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, Prove that (i) $$\triangle$$ ABC ~ $$\triangle$$ AMP (ii) $$CA\over PA$$ = $$BC\over MP$$

## Solution :

(i)  In triangles, ABC and AMP, we have

$$\angle$$ ABC = $$\angle$$ AMP = 90       (given)

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

$$\therefore$$  By AA similarity, we have

$$\triangle$$ ABC ~ $$\triangle$$ AMP

(ii)  We have :

$$\triangle$$ ABC ~ $$\triangle$$ AMP          (as proved above)

So, In similar triangles, corresponding sides are proportional.

$$\implies$$   $$CA\over PA$$ = $$BC\over MP$$