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  • On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that angle BAC equal to angle BDC.

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that \angleBAC = \angleBDC.

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

Let \triangleACB and \triangleADB are two right-angled triangles with common hypotenuse AB.

To prove: \angleBAC = \angleBDC

Proof:

Let O be the mid-point of AB

Then, OA = OB = OC = OD              

Mid-point of the hypotenuse of a right triangle is equidistant from both vertices.

We can draw a circle passing 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, \angleBAC and \angleBDC are angles of same segment BC.

\therefore \angleBAC = \angleBDC

Hence proved.

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