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

Given: AD ⊥ BC and CD > BD

To prove: AC > AB

Proof:

$$\begin{gathered} \angle ADB=\angle ADC=90°\left(AD\perp BC\right)...\left(1\right) \\ \angle BAD<\angle DAC\left(CD>BD\right)...\left(2\right) \end{gathered}$$

In $∆ABD$,

Using angle sum property of a triangle,

$$\begin{gathered} \angle B=180°-\angle ADB-\angle BAD \\ \angle B=90°-\angle BAD...\left(3\right) \end{gathered}$$

In $∆ADC$,

Using angle sum property of a triangle,

$\angle ACD=90°-\angle DAC...\left(4\right)$

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

$\angle B>\angle C$

Therefore, $AC>AB$.