In a triangle ABC, AB = AC and ∠A = 36°. If the internal bisector of ∠C meets AB at point D, prove that AD = BC.

In △ ABC,

⇒ AB = AC (Given)

∴ ∠B = ∠C = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 36° + x + x = 180°

⇒ 36° + 2x = 180°

⇒ 2x = 180° - 36°

⇒ 2x = 144°

⇒ x = $\dfrac{144°}{2}$ = 72°.

Since, CD is bisector of angle C,

∴ ∠ACD = ∠BCD = $\dfrac{72°}{2}$ = 36°.

In △ ACD,

⇒ ∠ACD = ∠DAC (Both equal to 36°)

∴ AD = DC [Sides opposite to equal angles are equal] ……..(1)

In △ DCB,

By angle sum property of triangle,

⇒ ∠CDB + ∠DCB + ∠DBC = 180°

⇒ ∠CDB + 36° + 72° = 180°

⇒ ∠CDB + 108° = 180°

⇒ ∠CDB = 180° - 108° = 72°.

∴ ∠CDB = ∠DBC (Both equal to 72°)

∴ BC = CD (Sides opposite to equal angles are equal) ………..(2)

From equation (1) and (2), we get :

⇒ AD = BC.

Hence, proved that AD = BC.