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We may prove theorems in two-dimensional projective geometry by using the freedom to project certain points in a diagram to, for example, points at infinity and then using ordinary Euclidean geometry to deal with the simplified picture we get.

**The harmonic property of the complete quadrangle/quadrilateral.**A complete quadrangle

*ABCD*is a set of 4 vertices together with the set of 6 lines joining them.

These define the three*diagonal points**PQR*.*The cross ratio*(*A*,*C*;*P*,*X*) = -1

**Proof**

Project te points*Q*and*R*to points at infinity. This gives the diagram on the right in which*abcd*is a parallelogram and so*ap*=*pc*.

Thus, since*x*is at infinity also, (*a*,*c*;*p*,*x*) = -1 and the result follows.

**Remark**The range (

*A*,*Y*,*D*,*R*) is also harmonic. This gives a method of constructing such harmonic ranges and is the starting point for a variety of geometric constructions.**Desargues theorem**

(First proved by Girard Desargues (1591 to 1661) In 1639)

*Two triangles in perspective from a point have corresponding sides meeting in a line.*That is, in the diagram shown in which the lines

*AA*',*BB*',*CC*are concurrent in a point*P*, the meets of*AB*and*A*'*B*', of*AC*and*A*'*C*' and of*CB*and*C*'*B*' are collinear.

**Proof**

Project the points*P*and*R*to infinity to get the diagram shown.

By similar triangles,*ac*is parallel to*a*'*c*' and so*e*too is on the line at infinity and so*p*,*q*,*r*are collinear and so too are*P*,*Q*,*R*.

**Remarks**- One my also prove this result in 3-dimensional space by observing that the line
*PQR*is the meet of the planes*ABC*and*A*'*B*'*C*'. The two dimensional case then follows by projection of this 3D figure into a plane. - This result fails in some projective planes (over finite fields).

- One my also prove this result in 3-dimensional space by observing that the line

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