$ABCD$ is a quadrilateral such that $A$ is the centre of the circle passing through $B, C$ and $D$. Prove that $\angle CBD + \angle CDB = \frac{1}{2} \angle BAD$.

(N/A) $ABCD$ is a quadrilateral such that $A$ is the centre of the circle passing through $B, C$ and $D$. We have to prove that $\angle CBD + \angle CDB = \frac{1}{2} \angle BAD$.
Join $AC$.
Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle:
Therefore,$\angle CAD = 2 \angle CBD$ $....(1)$
And $\angle BAC = 2 \angle CDB$ $....(2)$
Adding $(1)$ and $(2)$,we get:
$\angle CAD + \angle BAC = 2(\angle CBD + \angle CDB)$
$\Rightarrow \angle BAD = 2(\angle CBD + \angle CDB)$
Hence,$\angle CBD + \angle CDB = \frac{1}{2} \angle BAD$.