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}} $
"

AcademicMathematicsNCERTClass 10

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.

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

Advertisements