In the given figure. ∆ABC and ∆AMP are two right triangles, right

1.	In the given figure. ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ∆ABC ~ ∆AMP
Answer» Data: ∆ABC and ∆AMP are two right triangles, right angled at B and M respectvely. To Prove: i) ∆ABC ~ ∆AMP (ii) \(\frac{CA}{PA} = \frac{BC}{MP}\) (i) In ∆ABC and ∆AMP, ∠ABC = ∠AMP = 90° (data) ∠CAB = ∠MAP (common) ∴ ∠ACB = ∠MPA ∴ These are equiangular triangles. Similarity criterion for ∆ is A.A.A. ∴ ∆ABC ~ ∆AMP (ii) ∆ABC ~ ∆AMP(Proved) ∴ Corresponding sides are in proportion. LC and LP are corresponding angles. ∴ Adjacent sides are CA, PA. Similarly BC and MP are adjacent sides. ∴ \(\frac{CA}{PA} = \frac{BC}{MP}\)

In the given figure. ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ∆ABC ~ ∆AMP

Answer»

Data: ∆ABC and ∆AMP are two right triangles, right angled at B and M respectvely.

To Prove: i) ∆ABC ~ ∆AMP

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

(i) In ∆ABC and ∆AMP,

∠ABC = ∠AMP = 90° (data)

∠CAB = ∠MAP (common)

∴ ∠ACB = ∠MPA

∴ These are equiangular triangles.

Similarity criterion for ∆ is A.A.A.

∴ ∆ABC ~ ∆AMP

(ii) ∆ABC ~ ∆AMP(Proved)

∴ Corresponding sides are in proportion.

LC and LP are corresponding angles.

∴ Adjacent sides are CA, PA.

Similarly BC and MP are adjacent sides.

∴ \(\frac{CA}{PA} = \frac{BC}{MP}\)