Question

The arc of a circle subtending a right angle at any point of the circle in its alternate segment, is a semicircle.

Prove it!! ​

Answer 1

$\large\boxed{\underline{\mathcal{\red{S}\green{O}\pink{L}\orange{U}\blue{T}\red{I}\pink{O}\red{N:-}}}}$

GIVEN :–

Let an arcAB of a circle with centre 0 which subtends angle ∠ACB in arcBA such that ∠ACB = 90°.

TO PROVE :–

AB is a semicircle.

CONSTRUCTION:–

Join OA and OB.

PROOF :–

Since arcAB subtends ∠AOB at the centre and Z∠ACB at the remaining part of the circle. Therefore,

∠AOB = 2L∠ACB = 2 x 90° = 180°

•°• AO and OB are in the same straight line.

Hence, arcAB is a semicircle.

Answer 2

5 Marks Answer :

Given : An arc AB of circle(O,r) which subtends ∠ACB at a point C of BA other than A and B, such that ∠ACB = 90°.

To Prove : AB is a semicircle.

Construction : Join OA and OB.

Proof : AB subtends ∠AOB at the centre O and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB

=> ∠AOB = 2 * 90°

=> ∠AOB = 180°

⇒⇒ OA and OB are on the same straight line.

Hence, AB is  semi-circle.

Hope it helps!