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Topic:
More on the Kosnita point and the reflection triangle
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1
Last Post:
Apr 26, 2010 9:42 AM
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More on the Kosnita point and the reflection triangle
Posted:
Aug 8, 2003 2:12 PM
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In my note "On the Kosnita Point and the Reflection Triangle", Forum Geometricorum 3 (2003) pages 105-111, I studied the "reflection triangle" A'B'C' of a triangle ABC, where A' is the reflection of A in BC, and similarly B' and C' are defined. I proved that the circles B'C'A, C'A'B, A'B'C concur.
What is, may be, new: If Ua, Ub, Uc are the centers of these circles, then the lines AUa, BUb, CUc concur at...
... the nine-point center of triangle ABC.
Here is a PROOF.
Let Rc and Ra be the circumcenters of triangles ABC' and BCA', respectively, i. e. the reflections of the circumcenter O of triangle ABC in the sides AB and BC. (This is because the circumcenter of a reflected triangle is the reflection of the circumcenter of the original triangle.)
On the other hand, both circumcenters Rc and Ub lie on the perpendicular bisector of BC'. Let this perpendicular bisector meet BC at X and BC' at C" (i. e., C" is the midpoint of BC'). Analogously, the circumcenters Ra and Ub lie on the perpendicular bisector of BA'. Let this perpendicular bisector meet AB at Y and BA' at A" (i. e., A" is the midpoint of BA').
Triangles XBC" and YBA" are similar (equal angles: angle XBC" = 180ð - 2B = angle YBA", and angle XC"B = 90ð = angle YA"B). Hence,
BX : BY = BC" : BA".
But BC" = BC' / 2 = BC / 2 and BA" = BA' / 2 = BA / 2, so that we obtain
BX : BY = BC : BA,
and the line XY is parallel to CA.
It should be well-known that the reflection of the circumcenter of a triangle in a sideline is the reflection of the opposite vertex in the nine-point center. Thus, the points Rc and Ra are the reflections of C and A in the nine-point center N of triangle ABC. I. e., the lines ARa and CRc intersect at N, and we have RcRa || CA.
Together with XY || CA, we easily see that the intersections of corresponding sides of the triangles RaUbRc and ABC are collinear. By the Desargues theorem, the two triangles are perspective, and the lines ARa, BUb, CRc concur. Thus, the line BUb passes through the nine-point center N. Analogously for the lines AUa and CUc.
Darij Grinberg
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