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

(A) $1$. Since $A$ is the centre of the circle passing through $B, C$,and $D$,we have $AB = AC = AD = r$ (where $r$ is the radius of the circle).
$2$. In $\triangle ABC$,since $AB = AC$,the angles opposite to these sides are equal,so $\angle ABC = \angle ACB$.
$3$. In $\triangle ADC$,since $AD = AC$,the angles opposite to these sides are equal,so $\angle ADC = \angle ACD$.
$4$. In $\triangle BCD$,the sum of angles is $180^{\circ}$. Thus,$\angle CBD + \angle CDB + \angle BCD = 180^{\circ}$.
$5$. Note that $\angle BCD = \angle ACB + \angle ACD$. Substituting this into the sum,we get $\angle CBD + \angle CDB + \angle ACB + \angle ACD = 180^{\circ}$.
$6$. Since $\angle CBD = \angle ACB$ and $\angle CDB = \angle ACD$ is not necessarily true,let's use the property of the central angle. The angle subtended by arc $CD$ at the centre is $\angle CAD$ and at the circumference is $\angle CBD$. Thus,$\angle CAD = 2 \angle CBD$ is only true if $B$ is on the major arc.
$7$. Correct approach: In $\triangle ABD$,$AB=AD$,so $\angle ABD = \angle ADB = (180^{\circ} - \angle BAD)/2 = 90^{\circ} - \frac{1}{2} \angle BAD$.
$8$. In $\triangle BCD$,$\angle CBD + \angle CDB = 180^{\circ} - \angle BCD$. Since $A$ is the centre,$\angle BCD = \frac{1}{2} (360^{\circ} - \angle BAD) = 180^{\circ} - \frac{1}{2} \angle BAD$ (angle at circumference is half the reflex angle at centre).
$9$. Therefore,$\angle CBD + \angle CDB = 180^{\circ} - (180^{\circ} - \frac{1}{2} \angle BAD) = \frac{1}{2} \angle BAD$.