# A Triangle of Antreas Hatzipolakis

What if applet does not run? |

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

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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

By the construction, ΔABM is right with

AF = BF = FM.

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

∠AFM = ∠ABC

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.

Further

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:

Michel Cabart came up with a different proof:

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

- M is middle of AD thus belongs to EF
- 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.)

### References

- Jan Vonk,
__On the Nagel Line and a Prolific Polar Triangle__,*Forum Geometricorum*Volume 8 (2008) 183-196

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny