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.
\(\implies\) \(CA\over PA\) = \(BC\over MP\)