Start from a triangle t=(ABC) , a point D and a line e. Draw the intersection points A',B',C' on the sides of t of the line (h) through D, parallel to e. Join AA', BB', CC' and build the triangle t'=(A''B''C'') of these lines. Repeat the procedure by replacing t with t'. All these triangles are perspective to t on fixed lines f, g, h, k, which depend only on t, D and e.
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Desargues theorem (perspective triangles) ![[0_0]](Desargues_Many_files/Desargues_Many_0_0_0.png)
![[0_1]](Desargues_Many_files/Desargues_Many_0_0_1.png)
![[1_0]](Desargues_Many_files/Desargues_Many_0_1_0.png)
![[1_1]](Desargues_Many_files/Desargues_Many_0_1_1.png)