# In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:(i) $\triangle \mathrm{ABC} \sim \triangle \mathrm{AMP}$(ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$"

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Given:

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

To do:

We have to prove that

(i) $\triangle \mathrm{ABC} \sim \triangle \mathrm{AMP}$

(ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$

Solution:

(i) In $\triangle ABC$ and $\triangle AMP$,

$\angle B=\angle AMP=90^o$

$\angle A=\angle A$           (common)

Therefore, by AA criterion,

$\triangle ABC \sim \triangle AMP$

Hence proved.

(ii) 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.

Updated on 10-Oct-2022 13:21:13