The trouble of proof.

Proof: As shown in the figure, extend the intersection circle G of AG to P.

Ap is the diameter.

∴∠ACP = 90° (the circumferential angle relative to the diameter is a right angle)

∴∠GAC+∞∠p = 90 (the sum of the inner angles of the triangle is equal to 180).

∠∠B =∠P p (in the same circle, the relative angles of the circle and the same arc are equal)

∴∠GAC+∠B=90