From the following figure, prove that : AB > CD.

In △ ABC,

⇒ AB = AC (Given)

⇒ ∠C = ∠B = 70°

By angle sum property of triangle,

⇒ ∠BAC + ∠B + ∠C = 180°

⇒ ∠BAC + 70° + 70° = 180°

⇒ ∠BAC + 140° = 180°

⇒ ∠BAC = 180° - 140° = 40°.

In △ ABD,

By angle sum property of triangle,

⇒ ∠BAD + ∠B + ∠D = 180°

⇒ ∠BAD + 70° + 40° = 180°

⇒ ∠BAD + 110° = 180°

⇒ ∠BAD = 180° - 110° = 70°.

From figure,

⇒ ∠CAD = ∠BAD - ∠BAC = 70° - 40° = 30°.

In △ ACD,

Since, ∠CDA > ∠CAD

∴ AC > CD (If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.)

Since, AB = AC

∴ AB > CD.

Hence, proved that AB > CD.