Desargues' Theorem with parallel lines

Back to Geometry homepage

In the diagram above, the triangles \(\Delta ABC\) and \(\Delta DEF\) are in perspective from the point \(O\). This means that the lines \(AD\), \(BE\) and \(CF\) are all concurrent at \(O\).

The point \(E\) is chosen so that \(DE\) is parallel to \(AB\) and \(F\) is chosen so that \(EF\) is parallel to \(BC\). Desargues' theorem then says that \(DF\) is parallel to \(AC\). You can see this in the diagram above, since the angles \(\angle OAC\) and \(\angle ODF\) are equal, therefore \(AC\) and \(DF\) are parallel by Euclid Proposition I.28.

The converse to Desargues' theorem is also true: if \(AB\) is parallel to \(DE\), \(AC\) is parallel to \(DF\) and \(BC\) is parallel to \(EF\), then the two triangles \(\Delta ABC\) and \(\Delta DEF\) are in perspective from a point (i.e. \(AD\), \(BE\) and \(CF\) are concurrent at the point \(O\)).

You can click and drag on the points \(A\), \(B\), \(C\) and \(O\) to move them. The point \(D\) is constrained to lie on the line \(OA\), but you can click and drag to move it along this line.
As you move the points around, the triangles \(\Delta ABC\) and \(\Delta DEF\) are constrained to be in perspective from the point \(O\).

This is a special case of the more general Desargues' theorem where the intersection points \(P\), \(Q\) and \(R\) all lie on the line at infinity (click the link for an explanation).