If ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that ∠MAP = ∠BAC. Prove that
Question:
If ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively such that ∠MAP = ∠BAC. Prove that
(i) $\triangle \mathrm{ABC} \sim \triangle \mathrm{AMP}$
(ii) $\mathrm{CAPA}=\mathrm{BCMP}$
Solution:
(1) It is given that $\triangle A B C$ and $\triangle A M P$ are two right angle triangles.
Now, in $\triangle A B C$ and $\triangle A M P$, we have
$\angle M A P=\angle B A C \quad$ (Given)
$\angle A M P=\angle B=90^{\circ}$
$\triangle A B C \sim \triangle A M P$ (AA Similarity)
(2) $\triangle A B C \sim \triangle A M P$
So, $\frac{C A}{P A}=\frac{B C}{M P}$ (Corresponding sides are proportional)