$ABC$ and $ADC$ are two right triangles with a common hypotenuse $AC$. Prove that $\angle CAD = \angle CBD$.

(N/A) We are given two right-angled triangles,$\Delta ABC$ and $\Delta ADC$,which share a common hypotenuse $AC$.
Since $\angle ABC = 90^{\circ}$ and $\angle ADC = 90^{\circ}$,both triangles are right-angled at $B$ and $D$ respectively.
Consider a circle with $AC$ as its diameter. Since $\angle ABC = 90^{\circ}$ and $\angle ADC = 90^{\circ}$,the points $B$ and $D$ must lie on this circle (because the angle in a semicircle is a right angle).
Thus,$A, B, C,$ and $D$ are concyclic points.
Now,consider the chord $CD$. The angles $\angle CAD$ and $\angle CBD$ are angles subtended by the same chord $CD$ in the same segment of the circle.
According to the theorem that angles subtended by the same arc (or chord) in the same segment of a circle are equal,we have:
$\angle CAD = \angle CBD$.
Hence proved.