# 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 $$QP\parallel BC$$.

• Dumb question possibly, but I just want to clarify. You initially say $M$ is just between $B,C$, but then later act as if it is the midpoint of the segment and act as if the lines from $P,Q$ to $M$ are bisectors of their respective angles. I just want to make sure that these latter facts (midpoint, angle bisectors) are, indeed, given since your wording is vague in this respect. – Eevee Trainer Nov 16 '18 at 10:13

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, $$PQ\parallel BC$$.
• This might be a silly question but how did you reach the last conclusion? I mean from $$\frac{AQ}{QB} = \frac{AP}{PC}$$ to $PQ || BC$ – Sauhard Sharma Nov 16 '18 at 10:22
• @user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$. – Batominovski Nov 16 '18 at 10:57