Desargues’ Theorem(Desargues’ theorem)
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Q.
Draw three rays from a single point, and then draw two triangles. If the corresponding sides of the two triangles are extended, the intersection points of the corresponding sides lie on a single straight line.
Proof 1.
This is the most intuitive proof.
Suppose △ABC and △A’B’C’ are perspective from a point O, that is, AA’, BB’, and CC’ meet at a single point O. Let the intersections of the corresponding sides be P = AB ∩ A’B’, Q = BC ∩ B’C’, and R = CA ∩ C’A’. We need to show that P, Q, and R lie on a single straight line. Let the planes containing △ABC and △A’B’C’ be π and π’, respectively. P = AB ∩ A’B’, AB ∈ π, A’B’ ∈ π’. Therefore, P ∈ π ∩ π’. Similarly, we obtain Q ∈ π ∩ π’ and R ∈ π ∩ π’. Since π ∩ π’ = ℓ (where ℓ is a straight line), P, Q, and R lie on the same straight line ℓ. ■
Proof 2.
Place O at the origin and denote the position vectors of A, B, and C by A , B , and C , respectively. Therefore, the position vectors of A’, B’, and C’ can be written as λA , μB , and νC , respectively. Since P ∈ AB, P = (1 - s)A + sB , and since P ∈ A’B’, P = (1 - t)λA + tμB. Solving these equations gives s = μ(1 - λ) / (μ - λ). Therefore,
In coordinates with respect to ABC, P ~ (1 / (μ - λ)) (λ(μ - 1), μ(1 - λ), 0). Similarly, we obtain Q ~ (1 / (ν - μ)) (0, μ(ν - 1), ν(1 - μ)), R ~ (1 / (λ - ν)) (λ(1 - ν), 0, ν(λ - 1)). The fact that the three points P, Q, and R are collinear is equivalent to the determinant of these three coordinate vectors being 0, so let us show this.
The determinant of the three coordinate vectors is a constant multiple of the determinant above, and the determinant above is D = λ(μ−1)μ(ν−1)ν(λ−1) + μ(1−λ)ν(1−μ)λ(1−ν) = 0. Therefore, P, Q, and R lie on a single straight line. ■
Input: 2026.08.28 10:04