In the given figure, sides AB = AC. Show that AD > AB.

In △ABC,

⇒ ∠ABC = ∠ACB [As, AC = AB and angle opposite to equal sides are equal]

We know that,

In an isosceles triangle, the base angles are acute as only one obtuse angle can exist in a triangle.

∴ ∠ABC and ∠ACB are acute angles.

From figure,

DBC is a straight line.

⇒ ∠ABD + ∠ABC = 180° [Linear pair]

⇒ ∠ABD = 180° - ∠ABC

Since, ∠ABC is an acute angle thus, ∠ABD is an obtuse angle.

∴ In △ABD, ∠ABD is the largest angle.

∴ ∠ABD > ∠ADB

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

Hence, proved that AD > AB.