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/PA = BC/MP

Given, ABC and AMP are two right triangles, right angled at B and M respectively.


(i) In \(\triangle\) ABC and \(\triangle\) AMP, we have,
\(\angle\) CAB = \(\angle\) MAP (common angles)
\(\angle\) ABC = \(\angle\) AMP = 90° (each 90°)
\(\triangle\) ABC ~ \(\triangle\) AMP (AA similarity criterion)


(ii) As, \(\triangle\) ABC ~ \(\triangle\) AMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence, CA/PA = BC/MP