A Triangle of Antreas Hatzipolakis

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In ΔABC, D, E, F are the midpoints of sides BC, AC, and AB, while X, Y, Z are the points of tangency of the incircle with those sides. Let M be the foot of the perpendicular from A to BI and N the foot of the perpendicular from A to CI. Then

  1. Points M, N, E, F are collinear,
  2. Points X, Y, M are collinear, as are the points X, Z, N.


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Copyright © 1996-2017 Alexander Bogomolny

By the construction, ΔABM is right with ∠AMB = 90°, implying

AF = BF = FM.

Hatzipolakis Triangle

It follows that ΔBFM is isosceles and ∠FBM = ∠FMB. Thus the sum of these angles equals ∠AFM. On the other hand, since BI bisects ∠ABC, the sum of equal angles FBM and CBM equals ∠ABC. We see that


so that FM||BC and, since F is the midpoint of AB, FM is a midline in ΔABC and passes through another midpoint, viz., E. This shows that M lies on EF. By symmetry the same holds for N.


 EM= FM - EF
  = AF - EF
  = AB/2 - BC/2
  = (AB - BC)/2
  = (AZ - CX)/2
  = (AY - CY)/2
  = (AC - 2·CY)/2
  = (2·EC - 2·CY)/2
  = EC - CY
  = EY,

implying that ΔMEY is isosceles: EM = EY. In addition, ∠XCE = ∠CEM wherefrom ∠XYC = ∠EYM making therm vertical and XYM a straight line. XZN is treated similarly.

Michel Cabart came up with a different proof:

Consider point D of intersection of lines AM and BC. Then

  1. M is middle of AD thus belongs to EF
  2. by symmetry XD = ZA = YA thus XC·YA·MD = XD·YC·MA and M belongs to line XY by Menelaus' theorem applied to triangle ACD.

(Triangle AMN along with two other triangles similarly constructed at vertices B and C have several engaging properties first observed by Antreas Hatzipolakis which explains the attribution in the title.)


  1. Jan Vonk, On the Nagel Line and a Prolific Polar Triangle, Forum Geometricorum Volume 8 (2008) 183-196

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Copyright © 1996-2017 Alexander Bogomolny


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