Question

D is a point on side BC of a Δ ABC such that AD = AC. Show that AB > AD

Answer 1

In the above figure, AD = AC ( given)

So, < d = < c

In triangle ABD

Exterior angle d > angle b ( as exterior angle is greater than its interior opposite angle.)

But < d = < c

=> <c greater than <b

=> AB > AC ( as, side opposite to greater angle is greater in a triangle)

But, AC = AD

So, AB > AD

[ Hence Proved]In the above figure, AD = AC ( given)

So, < d = < c

In triangle ABD

Exterior angle d > angle b ( as exterior angle is greater than its interior opposite angle.)

But < d = < c

=> <c greater than <b

=> AB > AC ( as, side opposite to greater angle is greater in a triangle)

But, AC = AD

So, AB > AD

[ Hence Proved]