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Difference between revisions of "Desargue's Theorem"
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~First draft of this page created by Enoch Yu. Please add more to improve this page!!
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Revision as of 05:51, 10 October 2025

Introduction

Desargues' Theorem is a renowned theorem that relates the axis and the center of perspective of two triangles. Therefore, before we dive into Desargues' Theorem, we need to have a gut feeling on Projective Geometry. As I mentioned, because we only need a ``gut feeling, it is not necessary to attain profound insights on Projective Geometry.

First and foremost, consider the two, not necessarily similar, triangles and a point $P$.

Desargues 1.png

Let's try and think of it like this. We have a light source $P$ that shines $\triangle{ABC}$. There, we get our outcome $\triangle{A'B'C'}$. Henceforth, we say that $A', B', C'$ are corresponding points of $A, B, C$ respectively. Similarly, we could state that $\overline{A'B'}, \overline{B'C'}, \overline{C'A'}$ are corresponding sides of $\overline{AB}, \overline{BC}, \overline{CA}$ respectively.

Desargues' Theorem

Two triangles are axially perspective $iff$ they are centrally perspective. Two triangles are considered axially perspective if there exists the axis of perspectivity and they are centrally perspective if there exists the center of perspectivity.

For simpler explanation, consider the diagram below, which is simply the extension of Figure 1.

Desargues 2.png

Point $P$, where $AA', BB', CC'$ concur, is the center of perspectivity. Line $XZ$, where $X, Y, Z$ are collinear, is the axis of perspectivity in the diagram above. If the pairs of corresponding lines are all parallel, it is said that the lines meet at a point of infinity. However, the case is not commonly discussed.

The theorem is stating that if there exists a point $P$ such that $P = AA' \cap BB' \cap CC'$, then the points $X = AB \cap A'B'$, $Y = BC \cap B'C'$, and $Z = CA \cap C'A'$ are collinear.

Proof of Desargues' Theorem

Two triangles are axially perspective $iff$ they are centrally perspective.

The proof involves the use of Menelaus' theorem three times. Recall the the Menelaus' Theorem could be defined as the following. 

Menelaus' Theorem
Assume for $\triangle{ABC}$, $X, Y, Z$ are defined as $X \in \overline{AB}$, $Y \in \overline{BC}$, and $Z \in \overline{CA}$. Then the following equation of cross-ratio holds.
\[\frac{XA}{XB} \cdot \frac{YB}{YC} \cdot \frac{ZC}{ZA} = -1\]

Because Desargues' Theorem must satisfy the if and only if condition, let's first prove that two triangles are axially perspective if they are centrally perspective.

Assuming that $\triangle{ABC}$ and $\triangle{A'B'C'}$ are perspective from point $P$, Menelaus' Theorem could be used.

Desargues 3.png

Let's multiply our equations. \begin{align*} \left( \frac{XA'}{XB'} \cdot \frac{BB'}{BP} \cdot \frac{AP}{AA'} \right) \cdot \left( \frac{ZB'}{ZC'} \cdot \frac{CC'}{CP} \cdot \frac{BP}{BB'} \right) \cdot \left( \frac{YC'}{YA'} \cdot \frac{CP}{CC'} \cdot \frac{AA'}{AP} \right) &= 1 \\ \frac{XA'}{XB'} \cdot \frac{ZB'}{ZC'} \cdot \frac{YC'}{YA'} &= 1 \end{align*} With the equation, consider the diagram below.

Desargues 4.png

By converse of Menelaus' Theorem, it is evident that $X$, $Y$, and $Z$ are collinear.

Now that the first statement for if and only if condition is proven, the reverse must also be proven. Meaning, the statement ``two triangles are centrally perspective if they are axially perspective must be proven. Consider the diagram below. The same construction method was used. However, the if $AA'$, $BB'$, and $CC'$ concur is yet to be known.

Desargues 5.png

Notice that $\triangle{XBB'}$ and $\triangle{YCC'}$ are in perspective from point $Z$. Moreover, the first half of the proof manifested that $P$, $A$, and $A'$ must be collinear. By construction, $P$ is on line $BB'$ and $CC'$. Moreover, it is also on the line $AA'$. Thus, $\triangle{ABC}$ and $\triangle{A'B'C'}$ are centrally perspective if they are axially perspective.

Problem

$D$ is a point on $\overline{BC}$ in $\triangle{ABC}$. Let $I_1$, $I_2$ be the incenter of $\triangle{ABD}$ and $\triangle{ACD}$ respectively. Moreover, let $I_3$ and $I_4$ be ex-centers in respect to $\angle{BAD}$ and $\angle{CAD}$ respectively. Show that $\overline{I_1 I_2}$, $\overline{I_3 I_4}$, and $\overline{BC}$ intersect at one point. (Source: Unknown)

$Proof.$ First, because $I_1$ and $I_2$ are the angle bisectors of $\angle{ABD}$ and $\angle{ACD}$ respectively, $I = BI_1 \cap CI_2$ is the incenter of $\triangle{ABC}$. Similarly, we know that $I' = BI_3 \cap CI_4$ is the ex-center of $\triangle{ABC}$ in respect that $\angle{BAC}$. Therefore $A$, $I$, and $I'$ are collinear.

With similar idea, we could prove that $A, I_1, I_3$ are collinear as well as $A, I_2, I_4$. In other words, $I_1I_3$ and $I_2I_4$ intersect at point $A$. Therefore, $BI_1 \cap CI_2$, $I_1I_3 \cap I_2I_4$, and $I_3B \cap I_4C$ are collinear in $\triangle{I_1BI_3}$ and $\triangle{I_2CI_4}$. By Desargues' theorem, $\overline{I_1 I_2}$, $\overline{I_3 I_4}$, and $\overline{BC}$ concurs. $\square$

You can check out much more detailed solution here: Solution :L)


~First draft of this page is created by Enoch Yu. Please add more to improve this page!!

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