In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate : (i) ∠CDE (ii) ∠DCE

Since, FAC is a straight line. ∴ ∠BAC + ∠FAB = 180° ⇒ ∠BAC + 128° = 180° ⇒ ∠BAC = 180° - 128° = 52°. In △ ABC, ⇒ ∠A = 52° ⇒ AB = AC (Given) ∴ ∠B = ∠C = x (let) [Angles opposite to equal sides are equal]. By angle, sum property of triangle, ⇒ ∠A + ∠B + ∠C = 180° ⇒ 52° + x + x = 180° ⇒ 52° + 2x = 180° ⇒ 2x = 180° - 52° ⇒ 2x = 128° ⇒ x = = 64° ⇒ ∠B = ∠C = 64°. In △ DBC, ⇒ BC = CD (Given) ⇒ ∠BDC = ∠DBC = 64° (Angles opposite to equal sides are equal) From figure, ⇒ ∠ADE = ∠ABC = 64° (Corresponding angles are equal) Since, ADB is a straight line. ∴ ∠ADE + ∠CDE + ∠BDC = 180° ⇒ 64° + ∠CDE + 64° = 180° ⇒ ∠CDE + 128° = 180° ⇒ ∠CDE = 180° - 128° = 52°. Hence, ∠CDE = 52°. (ii) Since, DE || BC ⇒ ∠DCB = ∠CDE = 52° (Alternate angles are equal) From figure, ⇒ ∠DCE = ∠ECB - ∠DCB = 64° - 52° = 12°. Hence, ∠DCE = 12°.