Question

If $\Delta ABD$ and$\Delta AMP$ are two right triangles right angled at B and M respectively such that $\angle MAP=\angle BAC$. prove that
(i) $\Delta ABC\sim \Delta AMP$
(ii) $\frac{CA}{PA}=\frac{BC}{MP}$

Answer 1

(i)
In ABC and AMP
ABC = AMP (right angles)
BAC= MAP (given)

180 - (ABC+BAC) = 180 - (AMP +MAP)
APM = ACB (proved)
Hence ABC $\sim$ AMP by AAA criteria.

(ii)
We proved that ABC AMP
Hence,
(Corresponding parts of similar triangles)
From the above equation,

Hence proved