16516Re: Simson lines

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• Jul 1 2:18 AM
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Dear Alexei
> The polar transformation and the properties of Simson lines allow to
prove next interesting fact.
> Let I be the incenter of ABC. For an arbitrary point P we denote as
P_A the point on the line PA such that IA is the bisector (internal or
external) of angle PIP_A. The points P_B, P_C are defined similarly.
> 1. If P_A, P_B, P_C are collinear then this line touches the incircle
of ABC.
> 2. The locus of points P with this property is the line.
> Can anybody prove this without using the polar transformation?

Here is a better presentation of a generalization :
Starting with a point J in the plane of ABC, the perpendicular lines
through J to JA, JB, JC intersect BC, CA, AB respectively at A', B', C'.
It is well known that A',B',C' are on a same line L(J).
For any point M,
Ma is the point of the line AM such as <MJA = <AJMa
Mb is the point of the line BM such as <MJB = <BJMb
Mc is the point of the line CM such as <MJC = <CJMc
Then Ma,Mb,Mc are collinear if and only if M lies on L(J)
In this case, the four lines A'Ma, B'Mb, C'Mc, MaMbMc touch the
inscribed conic with focus J
This can be proved with projective tools using the fact that M->Ma is
the harmonic homology with center A and axis the line JA'
* The harmonic homology (I don't know if this English name is correct)
with center O, axis L maps M to his harmonic conjugate wrt O and OM
inter L

I think that, if we consider three harmonic homologies with centers
A,B,C, the locus of M with three colinear images is generally a cubic,
even when the three axis concur. In the particular case of three axis
La,Lb,Lc going through a point J and such as there exists an involution
swapping (JA,La), (JB,Lb),(JC,Lc), then the cubic degenerates in three
lines :
the fixed lines of the involution and the line through La inter BC, Lb
inter CA, Lc inter AB
Friendly. Jean-Pierre
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