## Goormaghtigh and isogonals

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• Dear Hyacinthists, I m writing to send you an elementary proof - not using Menelaos theorem nor projective methods - of the theorem you call Goormaghtigh
Message 1 of 1 , Jan 16, 2007
Dear Hyacinthists, I'm writing to send you an elementary proof - not
using Menelaos theorem nor projective methods - of the theorem you
call "Goormaghtigh theorem" (see Hyacinthos messages #8664, #8745, #9302).
I don't know which was the original proof of Goormaghtigh.

The theorem says:
If A', B', C' are the points where the reflections of lines PA, PB, PC
in a
given line through P meet the sides BC, CA, AB then A', B', C' are
collinear.

If P is on the circumcircle the result is well known since it can be
easily proved tha P is the Miquel point of points A', B' C' with respect
triangle ABC. In fact, using directed angles, we have:
from ang. A'BC' = ang. CBA = ang. CPA = ang. A'PC' follows that P,
A', B, C'
lie on a circle,
from ang. B'CA' = ang. ACB = ang. APB = ang. B'PA' follows that P,
A', B',
C lie on a circle,
from ang. CA'B' = ang. CPB' = ang. BPC' = ang. BA'C' follows that A',
B', C'
lie on a line.
So in this case A'B'C' is the "generalized" Simson line of P wrt ABC.

If P is not on the circumcircle then call Q its isogonal conjugate wrt
triangle ABC and Q' its isogonal conjugate wrt triangle A B'C'.
We want to prove that Q=Q':
ang. BQC = ang. BAC + ang. CPB (since P and Q are isogonal
conjugates in triangle ABC)
= ang. C'AB'+ ang .B'PC' =
= ang. C'Q'B' (since P and Q are isogonal conjugates in
triangle AB'C')

Now I will use the result that if P and Q are isogonal conjugates wrt
to a triangle ABC and X, Y, Z are the reflections of P in the sides of
ABC then Q is the circumcenter of the XYZ triangle (for a proof see
Darij Grinberg's paper on isogonals in his site:
de.geocities.com/darij_grinberg/).

If X, Y, Z, are the reflections of P in the sides BC, CA, AB of
triangle ABC
and X' its reflection in the side B'C' of triangle AB'C' then:
ang. ZXY = ang. BQC (because QC is orthogonal to XY, QB is
orthogonal to XZ)
ang. ZX'Y = ang. C'Q'B' (because Q'B' is orthogonal to X'Y, Q'C' is
orthogonal X'Z)
so, since ang. C'Q'B' = ang. BQC, we get ang. ZXY = ang. ZX'Y.
From this follows that X, Y, Z, X' are concyclic,
but the center of the XYZ-circle is Q while the center of the
X'YZ-circle is
Q' so Q and Q' are the same point.

In a similar way it is found that Q is the isogonal point of P also wrt
triangles A'BC' and A'B'C.
Now, since Q is the isogonal conjugate of P with respect to triangle
A'BC'
and triangle AB'C' we have:
ang. BC'A' = ang. AC'A' = ang. AC'P + ang. PC'Q + ang. QC'A' = ang.
QC'B' +
ang.PC'Q + ang.BC'P = ang.BC'B' = ang. AC'B'; this means that A', B',
C' are
on a line. QED.

It can also be easily seen just from elementary reasoning that A'B'C'
envelopes an inconic whose foci are P and Q as the "reflection line"
through
P varies, this conic is an
ellipse if P is inside the XYZ-circle, an hyperbola if it is outside (a
parabola if P is an infinite point or on the circumcircle).
A sketch of proof is: if T is the point where QX' meets B'C' then PT=TX'
because B'C' is the perp. bisector of PX',
so, if P is inside the XYZ-circle, QT+PT=QT+TX'=QX'=cost. -> the locus
of T
is an ellipse and B'C' is the tangent in T (B'T' is the bisector of angle
PTX'); if P is outside...

The idea of using the fact that triangles ABC AB'C', A'BC', A'B'C
share the
same isogonal conj. point of P to prove this theorem is due to Gerhard
Hessenberg.
I've found it in his book "Grundlagen der Geometrie", published in 1930.
His proof is not elementary because it uses the "Calculus of
Reflections" (an approach, similar to analytic geometry, based on
group-theoretic calculations), here I've just tried to "translate" his
idea into the language of elementary geometry and directed angles.

Ciao!
Marcello Tarquini
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