In the figure, AB = AC and D is any point on BC, show that AB > AD.

In △ ABD and △ ACD,

⇒ AB = AC (Given)

⇒ ∠ABC = ∠ACB = θ (let) [Angles opposite to equal sides are equal]

⇒ AD = AD (Common side)

∴ △ ABD ≅ △ ACD (By S.A.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ ∠ADB = ∠ADC = x (let)

From figure,

⇒ ∠ADB + ∠ADC = 180°

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°.

∴ ∠ADB = ∠ADC = 90°.

We know that,

The base angles of isosceles triangle are acute angles.

∴ θ < 90°

In △ ABD,

⇒ ∠ADB > ∠ABD

⇒ AB > AD. [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.]

Hence, proved that AB > AD.