$AB$ and $AC$ are two chords of a circle with centre $P$. If the bisector of $\angle BAC$ passes through the centre $P$,prove that $AB = AC$.

(A) Let the bisector of $\angle BAC$ be $AD$,where $D$ is a point on the circle such that $P$ lies on $AD$.
In $\triangle APB$ and $\triangle APC$:
$1$. $AP = AP$ (Common side).
$2$. $\angle BAP = \angle CAP$ (Since $AP$ is the bisector of $\angle BAC$).
$3$. $PB = PC$ (Radii of the same circle).
However,we need to show $AB = AC$.
Consider the triangles $\triangle APB$ and $\triangle APC$. Since $P$ is the centre,$PB = PC = r$.
Since $AP$ is the bisector,$\angle BAP = \angle CAP = \theta$.
Using the Law of Sines in $\triangle APB$: $\frac{PB}{\sin \theta} = \frac{AB}{\sin \angle APB} \implies AB = \frac{r \sin \angle APB}{\sin \theta}$.
Using the Law of Sines in $\triangle APC$: $\frac{PC}{\sin \theta} = \frac{AC}{\sin \angle APC} \implies AC = \frac{r \sin \angle APC}{\sin \theta}$.
Since $A, P, D$ are collinear,$\angle APB + \angle APC = 180^{\circ}$.
Thus,$\sin \angle APB = \sin(180^{\circ} - \angle APC) = \sin \angle APC$.
Therefore,$AB = AC$.