In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
$ \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}} $
"
Given:
ABC and AMP are two right triangles, right angled at B and M respectively.
To do:
We have to prove that \( \triangle \mathrm{ABC} \sim \triangle \mathrm{AMP} \)
Solution:
In $\triangle ABC$ and $\triangle AMP$,
$\angle B=\angle AMP=90^o$
$\angle A=\angle A$ (common)
Therefore, by AA criterion,
$\triangle ABC \sim \triangle AMP$
This implies,
$\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$ (CPCT)
Hence proved.
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