The vertex of angle $\angle BAC$ lies inside of a circle. Prove that the value of angle $\angle BAC$ is equal to half the sum of angle measures of the arcs of the circle confined inside angle itself and inside the angle symmetric to it through vertex $A$.
So I don't understand how to prove this. I've already drawn a diagram but I can't figure out how to prove this.
Please help! Thank you!
Best Answer
Let $BD$ and $CE$ be chords of the circle and $BD\cap CE=\{A\}$.
Thus, $$\measuredangle BAC=\measuredangle BEA+\measuredangle EBA=\frac{1}{2}\hat{BC}+\frac{1}{2}\hat{DE}.$$