In the given figure, AD ⊥ BC and CD > BD. Show that AC > AB.

Given: AD ⊥ BC and CD > BD

To prove: AC > AB

Proof:

$\angle A D B=\angle A D C=90^{\circ} \quad(A D \perp B C) \quad \ldots(1)$

$\angle B A D<\angle D A C \quad(C D>B D) \quad \ldots(2)$

In $\Delta A B D$

Using angle sum property of a triangle,

$\angle B=180^{\circ}-\angle A D B-\angle B A D$

$\angle B=90^{\circ}-\angle B A D \quad \ldots(3)$

In $\Delta A D C$

Using angle sum property of a triangle,

$\angle A C D=90^{\circ}-\angle D A C \quad \ldots(4)$

From (2), (3) and (4), we get

$\angle B>\angle C$

Therefore, $A C>A B$.