The vertex of angle BAC lies inside of a circle. Prove that the value of 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.

The vertex of angle $\mathrm{\angle }BAC$ lies inside of a circle. Prove that the value of angle $\mathrm{\angle }BAC$ 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!
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Jean Deleon
Let $BD$ and $CE$ be chords of the circle and $BD\cap CE=\left\{A\right\}$.
Thus,
$\measuredangle BAC=\measuredangle BEA+\measuredangle EBA=\frac{1}{2}\stackrel{^}{BC}+\frac{1}{2}\stackrel{^}{DE}.$