Question

Question 8
On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. Prove that $\angle BAC=\angle BDC$

Answer 1

Given, $\Delta ACBand\Delta ADB$ are two right-angled triangles with common hypotenuse AB.

To prove that $\angle BAC=\angle BDC$
Construction : Join CD

Proof
Let O be the mid-point of AB.
Then, OA = OB = OC = OD
Since mid-point of the hypotenuse of a right angled 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 OA.
we know that angles in the same segment of a circle are equal.
From the figure, $\angle BAC=\angle BDC$ are angles in same segment BC.
$\therefore \angle BAC=\angle BDC$