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

(N/A) Given: In $\triangle ABD$,$\angle A = 90^{\circ}$ and $AC \perp BD$.
To prove: $AD^{2} = BD \cdot CD$.
Proof: Consider $\triangle ACD$ and $\triangle BCA$.
In $\triangle ACD$ and $\triangle BCA$:
$\angle ACD = \angle BCA = 90^{\circ}$ (Since $AC \perp BD$ and $\angle A = 90^{\circ}$).
$\angle CAD = \angle CBA$ (Since both are complementary to $\angle BAD$ or $\angle BDA$).
Therefore,by $AA$ similarity criterion,$\triangle ACD \sim \triangle BCA$.
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{CD}{AC} = \frac{AC}{BC} = \frac{AD}{AB}$.
Wait,let us consider $\triangle ACD$ and $\triangle BAD$.
In $\triangle ACD$ and $\triangle BAD$:
$\angle ACD = \angle BAD = 90^{\circ}$.
$\angle D = \angle D$ (Common angle).
Therefore,by $AA$ similarity criterion,$\triangle ACD \sim \triangle BAD$.
Since the triangles are similar,the ratio of their corresponding sides is equal:
$\frac{AC}{BA} = \frac{CD}{AD} = \frac{AD}{BD}$.
Taking the last two parts of the ratio:
$\frac{CD}{AD} = \frac{AD}{BD}$.
Cross-multiplying gives:
$AD^{2} = BD \cdot CD$.
Hence proved.