In the figure (2) given below, AB = AC. Prove that AB > CD.

Since, AB = AC.

∴ ∠ABC = ∠ACB = 70° (As angles opposite to equal sides of an isosceles triangle are equal.)

From figure,

⇒ ∠ACB + ∠ACD = 180°

⇒ 70° + ∠ACD = 180°

⇒ ∠ACD = 110°.

In △ACD,

⇒ ∠CAD + ∠ADC + ∠ACD = 180°

⇒ ∠CAD + 40° + 110° = 180°

⇒ ∠CAD + 150° = 180°

⇒ ∠CAD = 30°.

In △ACD,

∠ADC = 40°

∠CAD = 30°

∴ ∠ADC > ∠CAD

∴ AC > CD (As side opposite to greater angle is greater.)

Since, AB = AC,

∴ AB > CD.

Hence, proved that AB > CD.