In the figure alongside,

AB = AC

∠A = 48° and

∠ACD = 18°.

Show that : BC = CD.

In △ ABC,

⇒ AB = AC (Given)

⇒ ∠C = ∠B = x (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 48° + x + x = 180°

⇒ 48° + 2x = 180°

⇒ 2x = 180° - 48°

⇒ 2x = 132°

⇒ x = $\dfrac{132°}{2}$ = 66°.

∴ ∠B = ∠C = 66°.

From figure,

⇒ ∠DCB = ∠C - ∠ACD = 66° - 18° = 48°.

In △ BDC,

By angle sum property of triangle,

⇒ ∠BDC + ∠DCB + ∠CBD = 180°

⇒ ∠BDC + 48° + 66° = 180°

⇒ ∠BDC + 114° = 180°

⇒ ∠BDC = 180° - 114° = 66°.

Since, ∠BDC = ∠CBD

∴ BC = CD (Sides opposite to equal angles are equal).

Hence, proved that BC = CD.