Draw a Desargues figure in this way (the order of construction is chosen for in order to control the location of the perspective center and the perspective line).

• Construct triangle ABC with 3 lines as sides.
• Construct line DE and color it and line AB red.
• Construct point O as the intersection of lines AD and BE.
• Construct point F as a point on the line OC.
• Construct line EF and color it and line BC blue.
• Construct line FD and color it and line CA green.
• Construct the interior of triangle ABC and triangle DEF.
• Intersect the two red lines, the two blue lines and the two green lines to get points P, Q, R. They should be collinear. Construct line PQ and make this line thick; it is the perspective line p.

• Now you can experiment. Drag the vertices of the triangles so that the sides are approximately parallel. You will get a Thales figure as the perspective line move towards infinity..
• Then drag the vertices so that lines AD and BE are approximately parallel. Then O moves to a point at infinity.
• Now move the triangles so that DEF is approximately a translation of ABC. Where are O and p?
• If O is back on the screen, is it possible to move the triangles so that line p passes through O?

We start with an experiment showing some of the remarkable properties of the polar line of a point. Remember the polar line of P is the line through the inversion P' of P which is also perpendicular to line OP. (O is the center of the inverting circle.)

• Draw a circle with center O and also draw a point P outside the circle.
• Draw two secant lines through P intersecting the circle at AB and CD.
• Construct point E as the intersection of line AD and line BC. Construct point F as the intersection of the line AC and line BD.
• Construct line EF and construct the points S and T as the intersections of this line with the circle. Observe that lines PS and PT appear to be tangents to the circle.
• Also observe that line EF is perpendicular to line OP and passes through the inversion P' of P in the circle. Thus line EF is the polar of P with respect to the circle.

Move P inside the circle. Of course points S and T will disappear, but line EF should still be through P' and perpendicular to OP.

Drawn a circle with center A through B and then construct 6 points P1, P2, ,,,,m P6 on th the circle. Construct LINES (not segments) P1P2, P2P3, etc to form the extended sides h1, h2, h3, h4, h5, h6 of a hexagon.

Now intersect these pairs of lines: P1P2 and P4P5; P2P3 and P5P6, P3P4 and P6P1. Call these points X, Y, Z. Construct line XY. Z should also lie on this line. This is called the Pascal Line of the hexagon.

The important observation is that since any conic is the projection of a circle, this the fact that X, Y and Z are collinear holds also for any conic. This is Pascal's Theorem.

Constructing a conic.

In a new sketch place 5 points P1, … P5. Construct the lines h1 = P1P2, h2 = P2P3, h3 = P3P4, h4 = P4P5. Then draw a line h5 = P5L, where L is any point.

Important. WE DO NOT KNOW P6. But if we assume that h5 is a side of a hexagon inscribed in a conic, we can figure out what the Pascal line is and then we can construct P6 and also h6 = P6P1.

See if you can figure out how to do this.

Reward: If you have constructed P6, then construct a random circle centered at P5. Animate point M around P5 and trace P6 (or construct a locus) and you will see the conice through the 5 points P1, … P5!