My diagram 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 $QP\parallel BC$.

  • $\begingroup$ 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. $\endgroup$ – 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, $PQ\parallel BC$.

  • $\begingroup$ This might be a silly question but how did you reach the last conclusion? I mean from $$\frac{AQ}{QB} = \frac{AP}{PC}$$ to $PQ || BC$ $\endgroup$ – Sauhard Sharma Nov 16 '18 at 10:22
  • $\begingroup$ See mathwarehouse.com/geometry/similar/triangles/…. I used the converse of the Side-Splitter Theorem. Theorem 4.3 here also states and proves the converse: jwilson.coe.uga.edu/MATH7200/Sect4.1.html. $\endgroup$ – Batominovski Nov 16 '18 at 10:23
  • $\begingroup$ Thanks a lot Batominovski! But only a small problem! Did you draw a line from P to C ? Actually I don't see PC in my diagram! $\endgroup$ – user602338 Nov 16 '18 at 10:46
  • $\begingroup$ @user602338 What do you mean you don't see $PC$ in your diagram? In your diagram, you have $P$ on the side $AC$. $\endgroup$ – Batominovski Nov 16 '18 at 10:57
  • $\begingroup$ You have written PC in your proof above. Don't you? $\endgroup$ – user602338 Nov 16 '18 at 10:59

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