In the triangle $\triangle ABC$ , the point $M$ is between $B$ and $C$. And also the lines $MP$ and $MQ$ are bisectors of $\angle AMC$ and $\angle AMB$. It means that: $$\angle AMP=\angle PMC$$
$$\angle AMQ=\angle QMB$$ and $$BM=MC$$ So now the puzzle tells us to prove that:$$QP\parallel BC$$ So I know Thales's theorem and all relations between the similiar triangles. But I can't find any pairs of similiar triangles or any parallel lines to use the Thales's theorem! Please help me proving $QP\parallel BC$.
By the Angle Bisector Theorem, $$\frac{AQ}{QB}=\frac{AM}{MB}\text{ and }\frac{AP}{PC}=\frac{AM}{MC}\,.$$ Since $M$ is the midpoint of $BC$, we have $MB=MC$, whence $$\frac{AQ}{QB}=\frac{AP}{PC}\,.$$ Therefore, $PQ\parallel BC$.