Let $ABC$ be a triangle where $AC \geq AB$. Let $C_1$ be the midpoint of $AB$ and $B_1$ be the midpoint of $AC$. Let $D$ be the intersection of:

  1. The line perpendicular to the internal bisector of $C$ that passes through $C_1$.
  2. the line perpendicular to the internal bisector of $B$ passing through $B_1$.

Let the line passing through $D$ that is parallel to the internal bisector of $A$ intersect side $AC$ at $E$. Lastly, let $F$ be the intersection of the perpendicular bisector of $BC$ with the arc of the circumcircle of $ABC$ containing $A$. Show that $EF \perp AC$.

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I tried using Archimedes' theorem (regarding cleavers), but that does not give much. I also tried to consider the medial triangle of $ABC$ and its circumcircle. With it, Archimedes' theorem, we know that $DE$ is the angle bisector of one of the angle of the medial triangle of $ABC$.

I was able to reduce this to showing that DE passes through the midpoint of BC.

  • $\begingroup$ Where did you find the problem? $\endgroup$ – Sawarnik Jan 10 '15 at 19:10
  • $\begingroup$ added some progress $\endgroup$ – user198454 Jan 11 '15 at 5:03
  • $\begingroup$ And is there any particular reason you made a right triangle in your figure? Is it supposed to be a right triangle? $\endgroup$ – Sawarnik Jan 11 '15 at 8:38
  • $\begingroup$ no, there is no particular reason. I just did it like this so lines AI and DE would not be too close to each other. $\endgroup$ – user198454 Jan 11 '15 at 10:13
  • $\begingroup$ Try using homothety with respect to the centroid of $ABC$ and $-\frac{1}{2}$ as a coefficient and look at what happens to two exterior bisectors at points $B$ and $C$. $\endgroup$ – ElThor Jan 11 '15 at 17:57

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