In fig. 2.39 ABC and amp …

Given: In the figure, ABC and AMP are two right triangles, right angled at B and M respectively,

To prove:

1. {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}AMP
2. {tex}\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}{/tex}

Proof:

1. In {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}AMP
   {tex}\angle{/tex}ABC = {tex}\angle{/tex}AMP (1) ........ [Each equal to 90^o]
   {tex}\angle{/tex}BAC={tex}\angle{/tex}MAP (2).........[Common angle]
   In view of (1) and (2)
   {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}AMP ..........AA similarity criterion
2. {tex}\triangle {/tex}ABC {tex} \sim {/tex} {tex}\triangle {/tex}AMP.........Proved above in(i)
   {tex}\therefore {/tex} {tex}\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}{/tex}.........Corresponding sides of two similar triangles are proportional.