Desargues’ Theorem
Math 353
←−−→
Definition. Two triangles 4A1 B1 C1 and 4A2 B2 C2 are perspective from a point P if the lines A1 A2 ,
←−−→
←−→
B1 B2 , and C1 C2 are concurrent at P.
Figure 1.
A trilateral is a 3-sided polygon, i.e., a triangle with emphasis on its sides rather than its angles. Given
4ABC, let a denote the side opposite ∠A, let b denote the side opposite ∠B, and let c denote the side
opposite ∠C. Then 4abc denotes the trilateral with sides a, b, and c. Of course, 4ABC and 4abc denote
the same figure. Given line segments a and b, let ab denote the point at which the lines containing a and b
intersect.
Definition. Two trilaterals 4a1 b1 c1 and 4a2 b2 c2 are perspective from a line l if the points a1 a2 , b1 b2 , and
c1 c2 are collinear on l.
Figure 3.
1
Desargues’ Theorem. Two triangles in the real projective plane are perspective from a point if and only
if they are perspective from a line.
Proof. Suppose that triangles 4A1 B1 C1 and 4A2 B2 C2 are perspective from a point P as in Figure 3.
Then points P, A1 , and A2 are collinear on one line, points P, B1 , and B2 are collinear on another line,
and points P, C1 , and C2 are collinear on a third line. First suppose that 4A1 B1 C1 and 4A2 B2 C2 lie in
different planes Π1 and Π2 , respectively, as in Figure 3.
Figure 3.
Since P, A1 , B1 , A2 , and B2 are coplanar, the lines containing c1 and c2 meet at the point c1 c2 . But c1 ⊂ Π1
and c2 ⊂ Π2 , so c1 c2 lies on line l = Π1 ∩ Π2 . Likewise, b1 b2 and a1 a2 lie on l, and it follows that 4a1 b1 c1
and 4a2 b2 c2 are perspective from the line l (see Figure 4).
Figure 4.
2
Now suppose that 4A1 B1 C1 and 4A2 B2 C2 lie in the same plane Π and are perspective from point P . Let
P1 and P2 be any two points off the plane Π and collinear with P. Connect P1 and P2 to the vertices of
4A1 B1 C1 and 4A2 B2 C2 , respectively, as shown in Figure 5.
Figure 5.
←−→
←−→
Since the points P1 , P2 , A1 and A2 are coplanar, the lines P1 A1 and P2 A2 lie in this plane and meet at some
←−→
←−→
←−→
←−→
point A3 . Similarly, P1 B1 and P2 B2 meet at some point B3 , and P1 C1 and P2 C2 meet at some point C3
(see Figure 6).
Figure 6.
3
Then 4A1 B1 C1 and 4A3 B3 C3 are non-coplanar and perspective from P1 , and 4A2 B2 C2 and 4A3 B3 C3
are non-coplanar and perspective from P2 . Let Π3 be the plane of 4A3 B3 C3 . By the non-coplanar case
above, 4A1 B1 C1 and 4A3 B3 C3 are perspective from l = Π ∩ Π3 , and 4A2 B2 C2 and 4A3 B3 C3 are also
←−−→
←−−→
←−−→
←−−→
perspective from l. Consequently A1 B1 and A3 B3 are concurrent with l, and A2 B2 and A3 B3 are concurrent
←−−→ ←−−→ ←−−→
with l, in which case the four lines A1 B1 , A2 B2 , A3 B3 , and l are concurrent and the point c1 c2 lies on l.
←−−→ ←−−→ ←−−→
Likewise, the four lines A1 C1 , A2 C2 , A3 C3 , and l are concurrent so that b1 b2 lies on l, and the four lines
←−−→ ←−−→ ←−−→
B1 C1 , B2 C2 , B3 C3 , and l are concurrent so that a1 a2 lies on l. It follows that 4A1 B1 C1 and 4A2 B2 C2 are
perspective from l. The converse follows by duality. ¥
4