Problem 692

a) From a point $A$ lying outside a circle, rays $AB$ and $AC$ intersect this circle. Prove that the measure of angle $BAC$ is equal to half the difference of the angular measures of the arcs of the circle enclosed within this angle.

b) The vertex of angle $BAC$ is located inside the circle. Prove that the measure of angle $BAC$ is equal to half the sum of the angular measures of the arcs of the circle enclosed within angle $BAC$ and within the angle symmetric to it with respect to vertex $A$.