In the figure,$OA \cdot OB = OC \cdot OD$. Show that $\angle A = \angle C$ and $\angle B = \angle D$.

(N/A) Given: $OA \cdot OB = OC \cdot OD$
Rearranging the terms,we get:
$\frac{OA}{OC} = \frac{OD}{OB} \quad ...(1)$
Also,we have:
$\angle AOD = \angle COB \quad$ (Vertically opposite angles) $...(2)$
From $(1)$ and $(2)$,by the $SAS$ (Side-Angle-Side) similarity criterion,we have:
$\Delta AOD \sim \Delta COB$
Since the triangles are similar,their corresponding angles are equal:
Therefore,$\angle A = \angle C$ and $\angle D = \angle B$.