The median $ BM $ of a triangle $ ABC $ intersects the circumscribed circle at point $ K $. The circumcircle of the triangle $ KMC $ intersects the segment $ BC $ at point $ P $, and the circumcircle of $ AMK $ intersects the extension of $ BA $ at $ Q $. Prove that $ PQ> AC $.
Note that $BA\cdot BQ=BM\cdot BK=BC\cdot BP$ so the triangles $ABC$ and $PBQ$ are similar. Thus we need to show that
$$AC\le PQ\iff$$$$\iff AB\le BP=\frac{BM\cdot BK}{BC}\iff$$$$\iff AB\cdot BC\le BM\cdot \bigg(BM+\frac{AC^2}{4BM}\bigg)=\frac{2BC^2+2AB^2-AC^2}{4}+\frac{AC^2}{4}$$which is obvious.