Angle at the centre is double the angle at the circumference.

Statement:

Angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of the circle.

Given:

$\mathrm{\angle }\text{BOC}=$ Angle at the Center.

$\mathrm{\angle }\text{BAC}=$ Angle at the Circumference.

To Prove:

$\mathrm{\angle }3+\mathrm{\angle }4=2(\mathrm{\angle }1+\mathrm{\angle }2)$

    OR$\mathrm{\angle }\text{BOC}=2\mathrm{\angle }\text{BAC}$

Construction:

    Join $\text{AO}$ and extend it to point $\text{D}$.

Proof:

❮ ❯

In $\mathrm{\Delta }\text{AOB},$

$\text{OA}=\text{OB}$

(Radii of the circle.)

$\mathrm{\angle }1=\mathrm{\angle }\text{B}$ (1)

(Angles opposite to equal sides.)

$\mathrm{\angle }3=\mathrm{\angle }1+\mathrm{\angle }\text{B}$

(Exterior Angle Theorem)

$\mathrm{\angle }3=\mathrm{\angle }1+\mathrm{\angle }1$ [From (1)]

$\mathrm{\angle }3=2\mathrm{\angle }1$ (2)

In $\mathrm{\Delta }\text{AOC},$

$\text{OA}=\text{OC}$

(Radii of the circle.)

$\mathrm{\angle }2=\mathrm{\angle }\text{C}$ (3)

(Angles opposite to equal sides.)

$\mathrm{\angle }4=\mathrm{\angle }2+\mathrm{\angle }\text{C}$

(Exterior Angle Theorem)

$\mathrm{\angle }4=\mathrm{\angle }2+\mathrm{\angle }2$ [From (3)]

$\mathrm{\angle }4=2\mathrm{\angle }2$ (4)

From (2) and (4), we get

$\mathrm{\angle }3+\mathrm{\angle }4=2\mathrm{\angle }1+2\mathrm{\angle }2$ 

$\mathrm{\angle }3+\mathrm{\angle }4=2(\mathrm{\angle }1+\mathrm{\angle }2)$

$\boxed{{\displaystyle \mathrm{\angle }\text{BOC}=2\mathrm{\angle }\text{BAC}}}$

$$ Hence Proved.