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:
$\hspace{1.5 cm}\text{OA} = \text{OB}$
(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.