Question

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

Answer 1

Given: ΔACB and ΔADB are two right angled triangles with common hypotenuse AB.

To prove: ∠BAC = ∠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 triangle is equidistant from its verticles.

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, ∠BAC and ∠BDC are angles of same segment BC.

∴ ∠BAC = ∠BDC

Hence proved.