$AB$ and $AC$ are two equal chords of a circle. Prove that the bisector of the angle $BAC$ passes through the centre of the circle.

(N/A) To prove: The bisector $AM$ of $\angle BAC$ passes through the centre $O$.
Construction: Join $BC$. Let the bisector $AM$ intersect $BC$ at $P$.
Proof: In $\Delta BAP$ and $\Delta CAP$:
$AB = AC$ (Given,as chords are equal)
$\angle BAP = \angle CAP$ (Given,as $AM$ is the bisector)
$AP = AP$ (Common side)
Therefore,$\Delta BAP \cong \Delta CAP$ (By $SAS$ congruence rule).
Thus,$BP = CP$ and $\angle BPA = \angle CPA$ (by $CPCT$).
Since $\angle BPA + \angle CPA = 180^{\circ}$ (Linear pair),
$\angle BPA = \angle CPA = 90^{\circ}$.
This means $AP$ is the perpendicular bisector of the chord $BC$. We know that the perpendicular bisector of any chord of a circle always passes through the centre of the circle. Hence,the bisector of $\angle BAC$ passes through the centre $O$.