$\triangle ABD$ is a right triangle right-angled at A and $AC \perp BD$. Show that
$\frac{AB^2}{AC^2}=\frac{BD}{DC}$.

Given:

$\triangle ABD$ is a right triangle right-angled at A and $AC \perp BD$. 

To do: 

We have to show that $\frac{AB^2}{AC^2}=\frac{BD}{DC}$.

Solution:

In $\triangle ABD$ and $\triangle ABC$,

$\angle DAB=\angle ACB=90^o$

$\angle B=\angle B$   (Common angle)

Therefore,

$\triangle ADB \sim\ \triangle CAB$   (By AA similarity)

This implies,

$\frac{AB}{CB}=\frac{BD}{AB}$   (Corresponding parts of similar triangles are proportional)

$AB.AB=CB.BD$   (By cross multiplication)

$AB^2=BC.BD$......(i)

Let $\angle CAB=x$,

This implies,

$\angle CAD=90^o-x$

In $\triangle CAB$,

$\angle CAB+\angle BCA+\angle ABC=180^o$

$x+90^o+\angle ABC=180^o$

$\angle ABC=180^o-90^o-x=90^o-x$

In $\triangle CAD$,

$\angle CAD+\angle CDA+\angle ADC=180^o$

$90^-x+90^o+\angle ADC=180^o$

$\angle ADC=180^o-180^o+x=x$

Therefore,

In $\triangle CAB$ and $\triangle CAD$,

$\angle CAB=\angle ADC=x$

$\angle ABC=\angle CAD=90^o-x$ 

Therefore,

$\triangle CAB \sim\ \triangle CDA$   (By AA similarity)

This implies,

$\frac{AC}{DC}=\frac{BC}{AC}$   (Corresponding parts of similar triangles are proportional)

$AC.AC=CB.DC$   (By cross multiplication)

$AC^2=BC.DC$.....(ii)

Dividing equations (i) and (ii), we get,

$\frac{AB^2}{AC^2}=\frac{BC.BD}{BC.DC}$

$\frac{AB^2}{AC^2}=\frac{BD}{DC}$

Hence proved.