1
$\begingroup$

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 $QP\parallel BC$.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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, $PQ\parallel BC$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.