If $S$ is the circumcentre of $\triangle ABC$ prove that $\angle BSD=\angle BAC$ where $D$ is the midpoint of $BC$

Question

If $S$ is the circumcentre of $\triangle ABC$ prove that $\angle BSD=\angle BAC$ where $D$ is the midpoint of $BC$.

I have tried a lot of ways but I have not been able to prove this. It would be great if you could give me a hint to solve this question.

Answer 1

Hint:

• $DS$ is the bisector of $BC$, but also the bisector of $\angle BSC$.
• Because $S$ is the center of circumcircle of $\triangle ABC$, $|\angle BSC| = 2|\angle BAC|$, due to inscribed angle theorem.

I hope this helps $\ddot\smile$