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° = ⇒ 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.