** Given : **ABC and ADC are two right triangles with common hypotenuse AC.

**To Prove :** ∠CAD = ∠CBD

**Proof :** Let O be the mid-point of AC.

Then OA = OB = OC = OD

Mid point of the hypotenuse of a right triangle is equidistant from its vertices with O as centre and radius equal to OA, draw a circle to pass through A, B, C and D.

We know that angles in the same segment of a circle are equal.

Since, ∠CAD and ∠CBD are angles of the same segment.

Therefore, ∠CAD = ∠CBD. **Proved.**