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Let ${D,E,F}$ be the feet of the altitude from ${A,B,C}$ in a ${\triangle{ABC}}$. Prove that the perpendicular bisector of ${EF}$ also bisects ${BC}$.

enter image description here

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The only important thing is that $\square BCEF$ lies on a circle, since both $E$ and $F$ look at $\overline{BC}$ at right angle. But $\overline{BC}$ is a diameter while $\overline{EF}$ is just a line segment connecting two points on a circle. Now the perpendicular bisector to that would intersect any diameter in the center of the circle, i.e. the midpoint of the diameter.

enter image description here

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    $\begingroup$ +1. I was going to make a similar argument. I'll just give you my diagram instead. :) $\endgroup$ – Blue Feb 21 '15 at 5:32
  • $\begingroup$ looks nice -- what did you make it with? $\endgroup$ – r0fg1 Feb 21 '15 at 5:36
  • $\begingroup$ I used GeoGebra. $\endgroup$ – Blue Feb 21 '15 at 5:41

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