Question

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

Answer 1

We know that AB is the common hypotenuse of △ ACB and △ ADB

So we get

∠ACB = 90^o

∠BDC = 90^o

It can be written as

∠ACB + ∠BDC = 180^o

We know that the opposite angles of a quadrilateral ABCD are supplementary

So we get ABCD as a cyclic quadrilateral which means that a circle passes through the points A, C, B and D

From the figure we know that the angles in the same segment are equal

∠BAC = ∠BDC

Therefore, it is proved that ∠BAC = ∠BDC.