Inside the triangle $ABC$ choose the point $M$, and on the side $BC$ - the point $K$ in such a way that $MK || AB$. The circle passing through the points $M, \, \, K, \, \, C,$ crosses the side $AC$ for the second time at the point $N$, a circle passing through the points $M, \, \, N, \, \, A, $ crosses the side $AB$ for the second time at the point $Q$. Prove that $BM = KQ$. (Nagel Igor)