In the figure,$\angle ACB = 90^{\circ}$ and $CD \perp AB$. Prove that $\frac{BC^{2}}{AC^{2}} = \frac{BD}{AD}$.

(N/A) In $\Delta ACD$ and $\Delta ABC$:
$\angle ADC = \angle ACB = 90^{\circ}$
$\angle CAD = \angle CAB$ (Common angle)
Therefore,$\Delta ACD \sim \Delta ABC$ (by $AA$ similarity criterion).
This implies $\frac{AC}{AB} = \frac{AD}{AC}$,so $AC^{2} = AB \cdot AD$ $...(1)$
In $\Delta BCD$ and $\Delta BAC$:
$\angle BDC = \angle BCA = 90^{\circ}$
$\angle CBD = \angle ABC$ (Common angle)
Therefore,$\Delta BCD \sim \Delta BAC$ (by $AA$ similarity criterion).
This implies $\frac{BC}{BA} = \frac{BD}{BC}$,so $BC^{2} = BA \cdot BD$ $...(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{BC^{2}}{AC^{2}} = \frac{BA \cdot BD}{AB \cdot AD}$
Since $BA = AB$,we get:
$\frac{BC^{2}}{AC^{2}} = \frac{BD}{AD}$