Gauss + Miquel + Aubert [a. k. a. Steiner]

grobber

We have a quadrilateral $ABCD$ . Let $E$ be the point of intersection of the lines $AB$ and $CD$ . Let $F$ be the point of intersection of the lines $AD$ and $BC$ . Prove the following facts:

a) The midpoints of the segments $AC$ , $BD$ , $EF$ are collinear.
The line through these midpoints is called the Gauss line of the quadrilateral $ABCD$ .

b) The orthocenters of the triangles $ABF$ , $CDF$ , $BCE$ , $ADE$ are collinear on a line perpendicular to the Gauss line.
The line joining these orthocenters is known as the Aubert line, or, also, as the Steiner line of the quadrilateral $ABCD$ .

c) The circumcircles of the four triangles $ABF$ , $CDF$ , $BCE$ , $ADE$ intersect at a point $M$ .
This point $M$ is called the Miquel point of the quadrilateral $ABCD$ .
The circumcenters of these four triangles $ABF$ , $CDF$ , $BCE$ , $ADE$ lie on a circle which also passes through $M$ .

**Posted:** Sun Sep 07, 2003 3:51 pm

Lagrangia

a) you use Menelau's theorem

**Posted:** Mon Sep 08, 2003 2:18 am

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As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.

Lagrangia

For the Aubert line:

You use the following:

Lemma :

Let ABC be a triangle and A' the projection of A on BC, B' the projection of B on CA and C' the projection of C on AB. Then we have HA*HA'=HB*HB'=HC*HC', where H is the orthocenter.

Proof :
It is easy to see that (HB'A with HA'B) and (HA'C and HC'A) (all are triangles) are proportional (this is not the right word .. I am sure.. can't find the right word, but you will figure out what I am trying to say Mr. Green

) we have:

HA*HA'=HB*HB' and HA*HA'=HC*HC' so lemma proved.

Proof of Aubert line :
We know from Gauss that the middle points of the diagonals AC, BD , EF are on Gauss line, which we denote d (lets consider the points : O1, O2,O3).

We consider the circles C1,C2,C3 of diameters AC, BD and EF. We denote H as the orthocenter of ADE and let A' be the projection of A on DE , D' the projection of D on AE and E' the projection of E on AD. It is easy to see that A' is on C1, D' on C2 and E' on C3. Using the previous lemma we get that: HA*HA'=HD*HD'=HE*HE'.
These relations show that H has equal point powers on C1,C2 and C3.

Now we consider dij the radical axe of circles Ci,Cj (i,j belong to {1,2,3} and i<>j). From HA*HA'=HD*HD' we have that H is on d12, from HD*HD'=HE*HE' we have that H is on d23. because d12 is perpendicular on d and d23 is perp on d, and from H we can only draw one perpendicular on d, we have that d12 and d23 are one and the same line,
Then d12=d23=d13=d' which is perpendicular on d.

We showed that H is on d'. You do the same with the ortocenters of ABF, DCF and BEC.

So theorem proved! Mr. Green

cheers!

**Posted:** Mon Sep 08, 2003 2:34 am

_________________
As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.

Lagrangia

Miquel point:

Lets consider M the 2nd point of meeting of the circumcircles of triangles:
BCE and DFC.

Because we have m(CMF)=m(CDA) and m(CBA)=m(CME) we have that m(BAF)+m(BMF)=m(CMF)+m(BMC)+m(BAF)=m(ADE)+m(EAD)+m(AED)=180. So we get that ABMF can be inscriebed. you do the same with AEMD. So you get what we call Miquel point!

cheers!

**Posted:** Mon Sep 08, 2003 2:45 am

_________________
As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.

grobber

Cool! That's prwtty much what I did too.

**Posted:** Mon Sep 08, 2003 2:59 am

galois · Riemann Hypothesis Offline Joined: 11 Jul 2003 Posts: 396

my proofs are same xcept for the 1st one where i used vectors and came up with a pretty neat solution Mr. Green

.neway,here's a related problem on complete quadrangles
each diagonal of a complete quadrangle is cut harmonically by the other two.try this.it's nice
best regards

**Posted:** Mon Sep 08, 2003 7:19 am

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reality continues to ruin my life

sprmnt21 · Riemann Hypothesis Offline Joined: 30 Oct 2003 Posts: 276

**Posted:** Tue Nov 02, 2004 2:11 pm

jayme · Yang-Mills Theory Offline Joined: 20 Nov 2007 Posts: 640

Dear Mathlinkers,
a synthetic proof of the orthogonality of the Gauss and Steiner's line can be found on my website
http://perso.orange.fr/jl.ayme vol. 4 La droite de Gauss et la droite de Steiner
Sincerely
Jean-Louis

**Posted:** Fri Nov 07, 2008 3:31 am

April

Related paper: http://forumgeom.fau.edu/FG2004volume4/FG200405.pdf Mr. Green

**Posted:** Fri Nov 07, 2008 5:08 pm

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balan razvan · Riemann Hypothesis Offline Joined: 10 Feb 2008 Posts: 376

see also miquel's star that uses his point
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1105712782&t=188934
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1132671#1132671

**Posted:** Tue Apr 21, 2009 11:54 am

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