Prove that in $\triangle ABC$ ,$QP\parallel BC$ 
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, $PQ\parallel BC$.
