Point O is the incircle center of triangle ABC. Point m is symmetric to point O relative to line AC and belongs

Point O is the incircle center of triangle ABC. Point m is symmetric to point O relative to line AC and belongs to the circumscribed circle of the triangle. prove that the angle ABC is 60 degrees.

Since, by condition, OP = PH, then the right-angled triangles AOB and AHP are equal in two legs, then the angle OAP = OHP = X0.

Similarly, triangles COP and CHP are equal, then the angle OCP = НСР = Y0.

Point O is the center of the inscribed circle, then OA and OS are the bisectors of angles A and C of triangle ABC.

Then the angle BAC = 2 * X0, the angle BCA = 2 * Y0.

The inscribed angle ABC rests on the arc AHC = AH + CH = (Y + X) 0.

The sum of the interior angles of the triangle ABC is equal to:

ABC + 2 * X + 2 * Y = 180.

ABC + 2 * (X + Y) = 180.

ABC = (Y + X), then 2 * (X + Y) = 2 * ABC.

Then ABC + 2 * ABC = 180.

Angle ABC = 180/3 = 600, as required.