Suppose ABC is a isosceles triangle with AB = AC. Side BA is produced to D such that BA = AD. Prove that ∠BCD is a right angle.

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Solution

Given: AB = AC = AD

To prove: ∠C = 90°

Proof:

In ΔABC:

∠B = ∠ACB … (1) (Angles opposite to equal sides are equal)

In ΔADC:

∠D = ∠ACD … (2) (Angles opposite to equal sides are equal)

In ΔBDC:

∠B + ∠C + ∠D = 180° (Sum of the angles of a triangle)

∠ACB + ∠C + ∠ACD = 180° (From (1) and (2)

∠C + (∠ACB + ∠ACD) = 180°

∠C + ∠C = 180°

2∠C = 180°

∴ ∠C = 90°

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