Take any point P on OC, draw a vertical line from P to OA, and cross OA to E, that is, PE is the distance from P to OA.

Similarly, draw PF and OB at f, and PF is the distance from p to OB.

Obviously, PEO angle = PFO angle is a right angle.

α bisects angle α0, so angle EOP= angle α0.

OP=OP again, so the triangle POE congruent triangles POF is obtained from the corner, so PE=PF, that is, the distances from P to OA and P to OB are equal.