A point $ M$ is taken on the perpendicular bisector of the side $ AC$ of an acute-angled triangle $ ABC$ so that $ M$ and $ B$ are on the same side of $ AC$. If $ \angle BAC=\angle MCB$ and $ \angle ABC+\angle MBC=180^{\circ}$, find $ \angle BAC.$
Source: Ukraine 2005 grade 10
Tags: geometry, perpendicular bisector, geometry proposed
A point $ M$ is taken on the perpendicular bisector of the side $ AC$ of an acute-angled triangle $ ABC$ so that $ M$ and $ B$ are on the same side of $ AC$. If $ \angle BAC=\angle MCB$ and $ \angle ABC+\angle MBC=180^{\circ}$, find $ \angle BAC.$