Desargues' Theorem
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In the diagram above, the triangles ABC and DEF are in perspective from the point O. This means that the lines AD, BE and CF (drawn in orange) are all concurrent at O.
The points P, Q and R are the intersection points of the corresponding sides of the triangles ABC and DEF. The point P is defined as the intersection of AB and DE, the point Q is defined as the intersection of BC and EF and the point R is defined as the intersection of AC and DF.
Desargues' theorem then guarantees that the intersection points P,Q and R are collinear. Another way of stating this is that if the triangles are in perspective from a point (i.e. AD, BE and CF are concurrent) then they are in perspective from a line (i.e. the intersection points P, Q, R are collinear).
The converse to Desargues' theorem is also true: if the two triangles are in perspective from a line (i.e. the intersection points P, Q, R are collinear) then they are in perspective from a point (i.e. AD, BE and CF are concurrent).
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. Similarly, the point E is constrained to lie on the line OB and the point F is constrained to lie on the line OC.
As you move the points around, the triangles ABC and DEF are constrained to be in perspective from the point O. You can see that the intersection points P, Q, R are always collinear, as predicted by Desargues' theorem.
If the intersection points P, Q, R all lie on the line at infinity, then the three pairs of lines are parallel. You can click here to see a picture of this case.
