Question

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

Answer 1

Given that:

$O$ is the Circumcenter of the triangle$△ABC$ and $D$ is the mid-point of the base $BC$.

To Prove:

$\angle BOD=\angle A$

Proof

Join $OB$ and $OC$

From$△OBD$ and $△OCD$

$OD=OD$ (common)

$BD=DC$ ( $D$ is the midpoint of $BC$ )

$OB=OC$ (radius of the circle)

By $SSS$ congruence rule,

$△OBD\cong △OCD$

$\angle BOD=\angle COD$ (By $CPCT$)

Assume $\angle BOD=\angle COD=x$

We know that,

The angle subtended by an arc at the Centre of the circle is twice the angle subtended by it at any other point in the remaining part of the circle.

Therefore,

$2\angle BAC=\angle BOC$

$2\angle BAC=\angle BOD+\angle DOC$

$2\angle BAC=x+x$

$2\angle BAC=2x$

$\angle BAC=x$

$\angle BAC=\angle BOD$

Hence, proved $\angle BOD=\angle A$