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

Given: In a triangle ABC, AB = AC and ∠A = 36°. The internal bisector of angle C meets AB at D.

To prove: AD = BC

Proof: Since AB = AC, Δ ABC is an isosceles triangle.

⇒ ∠B = ∠C

Sum of all angles of the triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ 36° + ∠B + ∠B = 180°

⇒ 36° + 2∠B = 180°

⇒ 2∠B = 180° - 36°

⇒ 2∠B = 144°

⇒ ∠B = $\dfrac{144°}{2}$

⇒ ∠B = ∠C = 72°

Since CD is the angle bisector of ∠C,

⇒ ∠ACD = ∠BCD = $\dfrac{1}{2}$ x ∠C

= $\dfrac{1}{2}$ x 72°

= 36°

In Δ ACD and Δ ABC,

∠DAC = ∠DCA = 36°

AB = AC (Given)

∠CAB = ∠CAD (common angle)

⇒ AD = CD ……………(1)

By ASA criterion,

Δ ACD ≅ Δ ABC

By CPCT, AD = CD.

In Δ DCB, sum of all angles of the triangle is 180°.

⇒ ∠DCB + ∠DBC + ∠CDB = 180°

⇒ 36° + 72° + ∠CDB = 180°

⇒ 108° + ∠CDB = 180°

⇒ ∠CDB = 180° - 108°

⇒ ∠CDB = 72°

As, ∠CDB = ∠CBD = 72°

⇒ CB = CD …………..(2)

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

AD = BC

Hence, AD = BC.