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 = $\dfrac{128°}{2}$ = 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°.