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A Triangle of Antreas Hatzipolakis
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
Copyright © 1996-2010 Alexander Bogomolny
By the construction, ΔABM is right with
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. Further
implying that ΔMEY is isosceles: Michel Cabart came up with a different proof: Consider point D of intersection of lines AM and BC. Then
(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
Copyright © 1996-2010 Alexander Bogomolny |
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