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In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.

How can it be proved that the points E, D, F are colinear?

  • 1
    $\begingroup$ You cannot in general ! because it is true only for some special position of $BC$ $\endgroup$ – Emilio Novati Dec 3 '18 at 10:29
  • $\begingroup$ What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention $\endgroup$ – Ki Yoon Eum Dec 3 '18 at 11:23
  • 1
    $\begingroup$ And... I solve it actually using Menelaus' theorem! $\endgroup$ – Ki Yoon Eum Dec 3 '18 at 12:23
  • $\begingroup$ Can you add your solution (as an answer)? $\endgroup$ – Berci Dec 3 '18 at 19:32

Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.

  • $\begingroup$ So, how is the Simpson "line" a line? $\endgroup$ – Oscar Lanzi Dec 4 '18 at 13:55
  • $\begingroup$ The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here. $\endgroup$ – user10354138 Dec 4 '18 at 13:57

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