**Definition:** Two triangles $\triangle_1 = \triangle(A_1, B_1, C_1)$ and $\triangle_2 = \triangle(A_2, B_2, C_2)$ are *in perspective w.r.t. a point* $S$ if

\[

S = (A_1 A_2) \cap (B_1 B_2) \cap (C_1 C_2)

\]

The triangles are *in perspective w.r.t. a line* $\ell$ if

\begin{align*}

A’ = (B_1 C_1) \cap (B_2 C_2)\\

B’ = (A_1 C_1) \cap (A_2 C_2)\\

C’ = (A_1 B_1) \cap (A_2 B_2)\\

\end{align*}

lie on $\ell$.

**Theorem (Desargues): **Let $\triangle_1 = \triangle(A_1, B_1, C_1)$ and $\triangle_2 = \triangle(A_2, B_2, C_2)$ be two triangles. Then $\triangle_1$ and $\triangle_2$ are in perspective w.r.t. a point if and only if they are in perspective w.r.t. a line.

**Proof:**

$\Rightarrow$: We may assume that $A_1, A_2, S$ and $B_1, B_2, S$ and $C_1, C_2, S$ are in general position on their respective lines. By a Lemma from Lecture 3 we may choose representative vectors

\begin{align*}

A_1 &= \sqvector{a_1}, A_2 = \sqvector{a_2}, S = \sqvector{s} &\text{ such that } a_1+a_2+s=0\\

B_1 &= \sqvector{b_1}, B_2 = \sqvector{b_2} &\text{ such that } b_1+b_2+s=0\\

C_1 &= \sqvector{c_1}, C_2 = \sqvector{c_2} &\text{ such that } c_1+c_2+s=0

\end{align*}

We have: $a_1+a_2=b_1+b_2=c_1+c_2=-s$. The points of intersection of corresponding sides are:

\begin{align*}

(A_1 C_1) \cap (A_2 C_2) &= \sqvector{a_1-c_1} = \sqvector{c_2-a_2}\\

(A_1 B_1) \cap (A_2 B_2) &= \sqvector{b_1-a_1} = \sqvector{a_2-b_2}\\

(B_1 C_1) \cap (B_2 C_2) &= \sqvector{c_1-b_1} = \sqvector{b_2-c_2}

\end{align*}

These three points lie on one line, because

\[

(a_1-c_1)+(b_1-a_1)+(c_1-b_1)=0

\]

$\Leftarrow$: Starting with triangles $\triangle_1 = \triangle(A_1, B_1, C_1)$ and $\triangle_2 = \triangle(A_2, B_2, C_2)$ in perspective w.r.t. the line $\ell = (A_3 B_3) = (A_3 C_3) = (B_3 C_3)$ we observe that the triangles $\triangle(A_1, A_2, B_3)$ and $\triangle(B_1, B_2, A_3)$ are in perspective w.r.t. $C_3$. Then the forward direction implies that

\begin{align*}

C_1 &= (A_1 B_3) \cap (B_1 A_3) \\

C_2 &= (B_2 A_3) \cap (A_2 B_3) \\

S &= (A_1 A_2) \cap (B_1 B_2)

\end{align*}

are collinear. Hence the triangles $\triangle_1$ and $\triangle_2$ are in perspective w.r.t. $S$.

**3d picture:**

If we consider two triangles in $\RP^3$ in perspective w.r.t. a point $P$, then each triangle spans a plane and these planes intersect in a line. The intersection points of corresponding sides of the triangles have to lie on the intersection line of the planes. And hence the triangles are in perspective w.r.t. this line.

**Symmetry of the Desargues configuration:**

Consider 5 points $A_1, A_2, \dots, A_5$ in $\RP^3$ in general position. Further let $\ell_{ij} = (A_i A_j)$ be the ten lines spanned by these points and let $E_{ijk} = (A_i A_j A_k)$ be the ten planes spanned by these points. Every line is contained in three planes and every plane contains three lines. Now intersect this configuration with a plane $E$ different from $E_{ijk}$ and introduce the following labels:

\begin{align*}

p_{ij} &= \ell_{ij} \cap E\\

g_{ij} &= E_{klm} \cap E \text{ with } \{i, j, k, l, m\} = \{1, 2, 3, 4, 5\}

\end{align*}

This labelling shows that there are $\binom{5}{2}$ pairs of perspective triangles.

Thank you fariwesen for the nice post. I added some pictures, which I hadn’t prepared in advance – in particular, the 3d picture 🙂

Does $\Delta(A,B,C) = \Delta(B,C,A)$ hold?

What is $\overleftrightarrow{AA} = A A$, i.e. what happens to $S$ in the first definition if $A_1 = A_2$?

Concerning “does $\Delta(A,B,C) = \Delta(B,C,A)$ hold?”:

The ordering/labeling of the points is important, because a different ordering implies that you consider different lines connecting the vertices of the triangles and intersections of different extended edges.

Concerning $A_1 = A_2$:

In this case, Desargues theorem can be understood as a trivial statement. The line connecting $A_1$ and $A_2$ disappears and the intersection points $A_1 B_1 \cap A_2 B_2$ and $A_1 C_1 \cap A_2 B_2$ coincide. Therefore, the statement is always true (because two lines always intersect in one point and two points always lie on one line).

In the proof of desargues theorem: Why does the statement holds that the three points lie on a line sincetheir sum is zero?

edit: nevermind, they lie on a projective line…