In the following figure;

AC = CD; ∠BAD = 110° and ∠ACB = 74°.

Prove that : BC > CD.

Since, BCD is a straight line.

∴ ∠BCA + ∠ACD = 180°

⇒ 74° + ∠ACD = 180°

⇒ ∠ACD = 180° - 74° = 106°.

In △ ACD,

AC = CD (Given)

∴ ∠CAD = ∠CDA = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠ACD + ∠CAD + ∠CDA = 180°

⇒ 106° + x + x = 180°

⇒ 106° + 2x = 180°

⇒ 2x = 180° - 106°

⇒ 2x = 74°

⇒ x = $\dfrac{74°}{2}$ = 37°.

From figure,

⇒ ∠BAC = ∠BAD - ∠CAD = 110° - 37° = 73°.

In △ ABC,

By angle sum property of triangle,

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ 73° + ∠ABC + 74° = 180°

⇒ ∠ABC + 147° = 180°

⇒ ∠ABC = 180° - 147°

⇒ ∠ABC = 33°.

∴ ∠ACB > ∠BAC > ∠ABC

∴ AB > BC > AC [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.] …….(1)

Given,

AC = CD ……(2)

From equations (1) and (2), we get :

BC > CD.

Hence, proved that BC > CD.