$A B C D$ is a quadrilateral such that $A$ is the centre of the circle passing through $B, C$ and $D$. Prove that $\angle C B D+\angle C D B=\frac{1}{2} \angle B A D$.
In the given figure, ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D.
Join AC and BD.
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.
Here, arc CD subtends ∠CAD at the centre and ∠CBD at B on the circle.
∴ ∠CAD = 2∠CBD ...(1)
Also, arc CB subtends ∠CAB at the centre and ∠CDB at D on the circle.
∴ ∠CAB = 2∠CDB ...(2)
Adding (1) and (2), we get
∠CAD + ∠CAB = 2(∠CBD + ∠CDB)
⇒ ∠BAD = 2(∠CBD + ∠CDB)
$\Rightarrow \angle C B D+\angle C D B=\frac{1}{2} \angle B A D$
Hence, $\angle C B D+\angle C D B=\frac{1}{2} \angle B A D$.