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Topic: More on the Kosnita point and the reflection triangle
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Darij Grinberg

Posts: 150
Registered: 12/3/04
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

Triangles XBC" and YBA" are similar (equal angles:
angle XBC" = 180° - 2B = angle YBA", and
angle XC"B = 90° = 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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