Question

prove that angle substained by an arc of a cirle at the center is double the angle subtended by it at any point on the remaining part of the circle

Answer 1

solution :-

Given:

Arc AB. Point C on the circle is outside AB.

To prove:

∠AOB = 2 × ∠ACB

Construction:

Draw a line CO extended till point D.

Proof:

In ΔOAC in each of these figures,
∠AOD = ∠OAC + ∠OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles)
OA = OC (Radii of same circle)
Thus, ∠ OAC = ∠ OCA (Angles opposite equal sides of a triangle are equal)
∠AOD = ∠OAC + ∠OCA
⇒∠AOD = 2 × ∠OCA

Similarly, in ΔOBC, ∠BOD = 2 × ∠OCB
∠AOD = 2 × ∠OCA and ∠BOD = 2 × ∠OCB
⇒ ∠AOD + ∠BOD = 2 ∠OCA + 2 ∠OCB
∠AOD + ∠BOD = 2 × (∠OCA + ∠OCB)
or ∠AOB = 2 × ∠ACB

Hence, the theorem is proved.