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

Solution:

Given, two right triangles ACB and ADB lie on common hypotenuse AB are situated on opposite sides.

We have to prove that ∠BAC = ∠BDC.

Join CD

Let O be the midpoint of AB

We know that the midpoint of the hypotenuse of a right triangle is equidistant from its vertices.

So, OA = OB = OC = OD

Now draw a circle that passes through the points A, B, C and D with O as centre and radius equal to OA.

We know that the angles in the same segment of a circle are equal.

From the figure,

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

Therefore, ∠BAC = ∠BDC

✦ Try This: In Fig 10.4 if ∠ABC = 20°, then ∠AOC is equal to

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10

NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 8

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

Summary:

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

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