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

(N/A) Given: In $\triangle ABD$,$\angle BAD = 90^\circ$ and $AC \perp BD$.
To prove: $AB^2 = BC \cdot BD$.
Proof: Consider $\triangle BCA$ and $\triangle BAD$.
In $\triangle BCA$ and $\triangle BAD$:
$\angle BCA = \angle BAD = 90^\circ$ (Given $AC \perp BD$ and $\angle BAD = 90^\circ$)
$\angle B = \angle B$ (Common angle)
Therefore,by $AA$ similarity criterion,$\triangle BCA \sim \triangle BAD$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{BC}{BA} = \frac{BA}{BD}$
Cross-multiplying gives:
$BA \cdot BA = BC \cdot BD$
$AB^2 = BC \cdot BD$.
Hence proved.