On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.
Given, $\triangle A C B$ and $\triangle A D B$ are two right angled triangles with common hypotenuse $A B$.
To prove $\angle B A C=\angle B D C$
Construction Join $C D$.
Proof Let $O$ be the mid-point of $A B$.
Then, $O A=O B=O C=O D$.
Since, mid-point of the hypotenuse of a right triangle is equidistant
from its vertices. Now, draw a circle to pass through the points $A, B, C$ and $D$. with $O$ as centre and radius equal to $O A$.
and radius equal to $O A$.
We know that, angles in the same segment of a circle are equal. From the figure, $\angle B A C$ and $\angle B D C$ are angles of same segment $B C$.
$\therefore \quad \angle B A C=\angle B D C$
Hence proved.