Let $$M$$ be the midpoint of the side $$AC$$ of $$\triangle ABC$$. Let $$P\in AM$$ and $$Q\in CM$$ be such that $$PQ=\frac{AC}{2}$$. Let circle $$(ABQ)$$ intersect with $$BC$$ at $$X\not= B$$ and circle $$(BCP)$$ intersect with $$BA$$ at $$Y\not= B$$. If $$\angle BYM = 45^{\circ}$$, find $$\angle BXM$$ in degrees.