In the figure,$ABD$ is a triangle right-angled at $A$ and $AC \perp BD$. Show that $AC^{2} = BC \cdot DC$.

(N/A) Given: In $\triangle ABD$,$\angle A = 90^{\circ}$ and $AC \perp BD$.
To prove: $AC^{2} = BC \cdot DC$.
Proof:
In $\triangle BCA$ and $\triangle ACD$:
$\angle BCA = \angle ACD = 90^{\circ}$ (Since $AC \perp BD$)
$\angle CBA = 90^{\circ} - \angle CAB$ (In $\triangle ABC$,$\angle B + \angle CAB = 90^{\circ}$)
Also,$\angle CAD = 90^{\circ} - \angle CAB$ (Since $\angle A = 90^{\circ}$,$\angle CAD + \angle CAB = 90^{\circ}$)
Therefore,$\angle CBA = \angle CAD$.
By $AA$ similarity criterion,$\triangle BCA \sim \triangle ACD$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{BC}{AC} = \frac{AC}{DC}$
Cross-multiplying,we get:
$AC^{2} = BC \cdot DC$.
Hence proved.