In the figure (3) given below, AC = CD. Prove that BC < CD.

From figure,

⇒ ∠ACB + ∠ACD = 180°

⇒ 70° + ∠ACD = 180°

⇒ ∠ACD = 110°.

In △ACD,

AC = CD

∴ ∠CAD = ∠CDA (As angles opposite to equal sides are equal.)

Let ∠CAD = ∠CDA = x.

In △ACD,

⇒ ∠CAD + ∠CDA + ∠ACD = 180°

⇒ x + x + 110° = 180°

⇒ 2x + 110° = 180°

⇒ 2x = 70°

⇒ x = 35°.

⇒ ∠CAD = 35°.

From figure,

⇒ ∠CAD + ∠CAB = 70°

⇒ 35° + ∠CAB = 70°

⇒ ∠CAB = 35°

In △ABC,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ ∠ABC + 70° + 35° = 180°

⇒ ∠ABC + 105° = 180°

⇒ ∠ABC = 75°.

In △ABC,

∠ABC > ∠BAC

⇒ AC > BC (As side opposite to greater angle is greater)

∵ AC = CD

∴ CD > BC or BC < CD.

Hence, proved that BC < CD.