- View SourceDear 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