Points P and Q are marked on the lateral sides AB and AC of the isosceles triangle ABC

Points P and Q are marked on the lateral sides AB and AC of the isosceles triangle ABC, so that the angle PXB = angle QXC where X is the middle of the base of BC. prove that ВQ = CP

In triangles BPX and CQX, BX = CX, since point X is the middle of BC, angle PBX = QCX since ABC is an isosceles triangle, angle PXB = QXC by condition.

Then the triangles are equal in side and two adjacent angles. Then BP = CQ.

In triangles ВРС and CBQ, the side ВС is common, PBX = QCX, ВР = СQ, then triangles ВРС and CBQ are equal on two sides and yeah between them.

Then BQ = CP, as required.