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Question
From the given figure, prove that ΔABC ~ ΔEDF
Solution
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
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