$O$ is the circumcentre of the triangle $ABC$ and $D$ is the mid-point of the base $BC$. Prove that $\angle BOD = \angle A$.

(N/A) Given: $O$ is the circumcentre of $\Delta ABC$ and $D$ is the mid-point of $BC$,so $OD \perp BC$.
To prove: $\angle BOD = \angle A$.
Construction: Join $OB$ and $OC$.
Proof: In $\Delta OBD$ and $\Delta OCD$,we have:
$OB = OC$ (Radii of the same circumcircle)
$OD = OD$ (Common side)
$BD = CD$ ($D$ is the mid-point of $BC$)
Therefore,$\Delta OBD \cong \Delta OCD$ by $SSS$ congruence rule.
This implies $\angle BOD = \angle COD$ (by $CPCT$).
Thus,$\angle BOC = \angle BOD + \angle COD = 2\angle BOD$.
We know that 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,$\angle BOC = 2\angle BAC = 2\angle A$.
Equating the two expressions for $\angle BOC$:
$2\angle BOD = 2\angle A$
$\angle BOD = \angle A$.
Hence,proved.