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### Desargues' Theorem

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Date: 09/22/1999 at 08:53:50
From: Rezvan
Subject: Algebra, Desargues' Theorem

I'd like to find out more about Desargues' theorem and the ways to
prove it. I've searched the Internet and many books in my college's
library. I found some materials on the Internet but none in the
library. I'd be grateful if you could help me and also give me some
questions related to Desargues' theorem.

Thank you.
```

```
Date: 09/22/1999 at 10:07:29
From: Doctor Anthony
Subject: Re: Algebra, Desargues' Theorem

Three concurrent straight lines 0A, 0B and 0C are produced to A', B',
and C', respectively. Prove that the points of intersection of AB and
A'B', BC and B'C', and CA and C'A' are collinear.

This is Desargues' theorem. If BC, B'C' meet at P; CA, C'A' meet at Q;
and AB, A'B' meet at R; then we must prove P, Q and R are collinear.

Use the well-known result that if P, Q, and R are collinear, then the
two conditions that will be satisfied are:

k1.p + k2.q + k3.r = 0
k1 + k2 + k3 = 0

where p, q and r are position vectors respectively of P, Q, and R, and
k1, k2, and k3 are scalar constants.

Since OAA' are collinear we have (taking v as position vector of O)

v + k1.a + k1'.a' = 0
1 + k1 + k1'     = 0

Since OBB' are collinear we have

v + k2.b + k2'.b' = 0
1 + k2 + k2'      = 0

Since OCC' are collinear we have

v + k3.c + k3'.c' = 0
1 + k3 + k3'     = 0

From the second and third pairs of equations

k2.b - k3.c    k2'b' - k3'c'
----------- =  -------------
k2 - k3        k2' - k3'

and these must both represent the point P where BC intersects B'C' = p
and so

(k2-k3) x p = k2.b - k3.c

Similarly

(k3-k1) x q = k3.c - k1.a

(k1-k2) x r = k1.a - k2.b

adding these three equations we get

(k2-k3)p + (k3-k1)q + (k1-k2)r = 0   and of course
(k2-k3) + (k3-k1) + (k1-k2)   = 0

and this is the condition for  P, Q, and R to be collinear.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/22/1999 at 11:54:43
From: Doctor Floor
Subject: Re: Algebra, Desargues' Theorem

Hi, Rezvan:

In addition to what Dr. Anthony stated, the complete theorem says:

Let ABC and A'B'C' be two triangles, with sides abc and a'b'c'
respectively. Then the following conditions are equivalent:

(1) AA', BB', CC' are concurrent
(2) The points of intersection of a+a', b+b' and c+c' are
collinear.

Dr. Anthony proved, using calculations, (1) --> (2). One could prove
in the same way that (2) --> (1), but this can also be seen "by
duality." Let me explain:

In geometry, there is some symmetry in points and lines:

* two points define one line
* two lines define one point

This symmetry is found in all of Euclid's axioms (but not the parallel
postulate). We say that points and lines are dual.

Each geometric statement about lines and points has a dual version:

* A triangle is formed out of three non-collinear points.
* A trilateral is formed out of three non-concurrent lines.

* Side a is the line through vertices B and C.
* Vertex A is the intersection of sides b and c.

There are some other, more vague dual forms:

* In Cartesian coordinates a point (a,b) is given by two numbers
a, b
* and a line y = mx+n is given by two numbers m and n.

The symmetry in Euclid's axioms has a consequence: when a certain
statement is correct, then so is its dual.

Duality does NOT mean that if a certain condition holds in a figure,
then immediately the dual condition holds in the same figure. Three
lines that intersect in a point do not imply three collinear points in
the same figure (whatever meaning that could have). Normally the dual
version needs a new figure. So three intersecting lines in one figure
are dual to three collinear points in another figure.

Back to Desargues' theorem. Note that conditions (1) and (2) of
Desargues' theorem are dual:

1. The pairs AA', BB', and CC' of the vertices of two triangles define
three concurrent lines,

2. The pairs of sides aa', bb' and cc' of the side lines of two
triangles define three collinear points.

Desargues' theorem says that these two duals hold together in one
figure. If you prove (1) -> (2), then the dual statement (2) -> (1)
must also be true. So Dr. Anthony did in fact fully prove Desargues'
theorem.

Another way to 'prove' Desargues' theorem is by seeing the two
triangles as projection of a (double) pyramid. Triangles ABC and
A'B'C' are formed by two planes that intersect the pyramid. The point
of intersection of AA', BB', and CC' is the top of the pyramid. The
line connecting the intersections aa', bb', and cc' is the line of
intersection of the two planes.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/23/1999 at 10:02:14
From: Rezvan Moghimi
Subject: Thank you!

Dear Dr. Math:

I'd like to thank both Dr. Anthony and Dr. Floor for answering my
question. Their answers helped me to have a better understanding of

I have one more question about Desargues' theorem. Since I'm doing a
project on this subject, I need to have a question as my thesis
statement. I'd be grateful if you can help me to find a suitable
question (a question that can be solved using Desargues' theorem.)

Thank you.
```

```
Date: 09/26/1999 at 05:04:47
From: Doctor Floor
Subject: Re: Thank you!

Dear Rezvan,

Your question is difficult, but I would suggest something like:

"Given two triangles with sides a, b, c and a', b', c', under what
conditions do the points of intersection X of a and a', Y of b and b',
and Z of c and c' lie on one line?"

I hope that is suitable.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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