Thread: Desargues's Theorem

Two-nonparallel lines are drawn on a sheet f paper so that their theoretical intersectionn is somewhere off the paper. Through a point, P, selected on the part of the paper between the lines, construct the line that would, when sufficiently extended, pass through the intersection of the given lines.

Any guidance as to where to begin (I'm guessing triangles or collinear points need to be drawn before Desargues's Theorem or its converse can be utilized) would be appreciated!

Hi -

Desargues's Theorem says that if two triangles are drawn in perspective from a point, then the points of intersection of their corresponding pairs of sides are collinear.

To solve your problem, we need to end up with a diagram like the one attached. l and m are the two initial lines and P is the given point. Triangles ABP and FGH are the ones that will end up being in perspective from the point of intersection off the paper. The dotted line n is the one we need to find, and the line k represents the line joining pairs of corresponding sides.

To draw the diagram, draw the given lines l and m, and the point P. Then the line k (more or less anywhere); then the other points in alphabetical order. Start with A and B in arbitrary positions on the lines m and l. This fixes C, D and E. Then choose any convenient point for F. Then find G and finally H. Join PH to find the line n.

Hope that helps.
Grandad