Question

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

Answer 1

Given that:

$△ACB$ and $△ADB$ are situated on opposite sides of a common hypotenuse $AB$

To Prove

$\angle BAC=\angle BDC$

Proof

$△ACB$ and $△ADB$ are right angled triangles,

Then,

$$\begin{gathered} \angle C+\angle D=90°+90° \\ \angle C+\angle D=180° \end{gathered}$$

Therefore $ADBC$ is a cyclic quadrilateral.

Sum of opposite angles of a cyclic quadrilateral $=180°$

As we know,

Angles formed in the same segment of a circle are equal.

$\angle BAC$ and $\angle BDC$ lie in the same segment of chord $BC$ .

Hence, it is proved that $\angle BAC=\angle BDC$.