On a common hypotenuse $AB$,two right triangles $ACB$ and $ADB$ are situated on opposite sides. Prove that $\angle BAC = \angle BDC$.

(N/A) In right triangles $ACB$ and $ADB$,we have:
$\angle ACB = 90^{\circ}$ and $\angle ADB = 90^{\circ}$
Therefore,$\angle ACB + \angle ADB = 90^{\circ} + 90^{\circ} = 180^{\circ}$.
Since the sum of a pair of opposite angles of the quadrilateral $ADBC$ is $180^{\circ}$,it is a cyclic quadrilateral.
Now,consider the arc $BC$. The angles $\angle BAC$ and $\angle BDC$ are subtended by the same arc $BC$ in the same segment of the circle.
Since angles subtended by the same arc in the same segment of a circle are equal,we have $\angle BAC = \angle BDC$.