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:

$\hspace{0.5 cm}\angle \text{BOC}=$ Angle at the Center.

$\hspace{0.5 cm}\angle \text{BAC}=$ Angle at the Circumference.

To Prove:

$\hspace{0.75 cm}\angle 3 + \angle 4 = 2\; (\angle 1 + \angle 2)$

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

Construction:

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

Proof:

❮ ❯

In $\Delta \text{AOB},$

$\hspace{1.5 cm}\text{OA} = \text{OB}$

(Radii of the circle.)

$\hspace{1.5 cm}\angle 1 = \angle \text{B}\hspace{2.3 cm}$ (1)

(Angles opposite to equal sides.)

$\hspace{1.5 cm}\angle 3 = \angle 1 + \angle \text{B}$

(Exterior Angle Theorem)

$\hspace{1.5 cm}\angle 3 = \angle 1 + \angle 1\hspace{1 cm}$ [From (1)]

$\hspace{1.5 cm}\angle 3 = 2\;\angle 1\hspace{2 cm}$ (2)

In $\Delta \text{AOC},$

$\hspace{1.5 cm}\text{OA} = \text{OC}$

(Radii of the circle.)

$\hspace{1.5 cm}\angle 2 = \angle \text{C}\hspace{2.3 cm}$ (3)

(Angles opposite to equal sides.)

$\hspace{1.5 cm}\angle 4 = \angle 2 + \angle \text{C}$

(Exterior Angle Theorem)

$\hspace{1.5 cm}\angle 4 = \angle 2 + \angle 2\hspace{1 cm}$ [From (3)]

$\hspace{1.5 cm}\angle 4 = 2\;\angle 2\hspace{2 cm}$ (4)

From (2) and (4), we get

$\hspace{0.75 cm}\angle 3 + \angle 4 = 2\;\angle 1 + 2\;\angle 2$ 

$\hspace{0.75 cm}\angle 3 + \angle 4 = 2\; (\angle 1 + \angle 2)$

$\hspace{1 cm}\boxed{\angle \text{BOC} = 2\;\angle \text{BAC}}$

$\hspace{1.5 cm}$ Hence Proved.