Question

From the given figure, prove that ΔABC ~ ΔEDF

Answer 1

From the ΔABC, AB = AC

It is an isosceles triangle

Angles opposite to equal sides are equal

∴ ∠B = ∠C = 65°

∴ ∠B + ∠C = 65° + 65°

= 130°

We know that sum of three angles is a triangle = 180°

∠A + ∠B + ∠C = 180°

∠A + 130° = 180°

∠A = 180° – 130°

∠A = 50°

From ΔEDF, ∠E = 50°

∴ Sum of Remaining angles = 180° – 50° = 130°

DE = FD

∴ ∠D = ∠F

From ΔABC and ΔEDF

∴ ΔD = `130/2` = 65°

∠A = ∠E = 50°

∠B = ∠D = 65°

∠C = ∠F = 65°

∴ By AAA criteria ΔEDF ~ ΔABC