In the given figure, AB = AC; ∠A = 50° and ∠ACD = 15°. Show that BC = CD.

In △ACD,

By angle sum property of triangle,

⇒ ∠ACD + ∠CDA + ∠DAC = 180°

⇒ 15° + ∠CDA + 50° = 180°

⇒ ∠CDA + 65° = 180°

⇒ ∠CDA = 180° - 65°

⇒ ∠CDA = 115°.

From figure,

⇒ ∠CDA + ∠BDC = 180° (Linear pair)

⇒ 115° + ∠BDC = 180°

⇒ ∠BDC = 180° - 115°

⇒ ∠BDC = 65° …..(1)

In △ABC,

AB = BC

⇒ ∠ABC = ∠ACB = x° (Angles opposite to equal sides in a triangle are equal)

By angle sum property of triangle,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ x° + x° + 50° = 180°

⇒ 2x° = 180° - 50°

⇒ 2x° = 130°

⇒ x° = $\dfrac{130°}{2}$

⇒ x° = 65°

⇒ ∠ABC = ∠ACB = 65° ….(2)

From eq.(1) and (2), we have :

⇒ ∠BDC = ∠ABC = 65°

Since,

⇒ ∠DBC = ∠ABC

Thus,

⇒ ∠DBC = ∠BDC

Thus, in triangle DBC,

⇒ BC = CD (Sides opposite to equal angles in a triangle are equal)

Hence, proved that BC = CD.