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Suppose $ABC$ is an acute-angled triangle with $AB<AC$.Let $M$ be the midpoint of $BC$. Suppose $P$ is a point on side $AB$ such that, if $PC$ intersects the median $AM$ at E, then $AP=PE$.Prove that $AB=CE$.

I don't know how to start. Please give me an idea. Getting no fruitful thoughts ,I started using barycentric Coordinates.But the calculations seemed very tough and I failed to proceed. Please give me any idea to start

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  • $\begingroup$ I successfully bashed the problem with barycentric coordinates within 1 page!! $\endgroup$ – Sufaid Saleel Sep 11 '18 at 2:17
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Hint:   write Menelaus' theorem for triangle $\,\triangle PBC\,$ and transversal $\,AM\,$.

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Construct the //gm BECF.

enter image description here

From (1) AEM is a straight line; (2) E is a vertex of that //gm; (3) M is the midpoint of one of its diagonals; and (4) F is a vertex of that //gm and FM is a straight line, we can say that AEMF is a straight line.

The required result follows because all the green marked angles are equal.

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  • $\begingroup$ What does //gm mean? $\endgroup$ – Aqua Jul 23 '18 at 19:02
  • $\begingroup$ @Angle I guess it stands for Parallelogram? $\endgroup$ – Mythomorphic Jul 23 '18 at 19:13
  • $\begingroup$ Nice solution +1 $\endgroup$ – Aqua Jul 23 '18 at 19:25
  • $\begingroup$ @Mythomorphic That is right. $\endgroup$ – Mick Jul 24 '18 at 2:45

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