Note 1
[G] Projective Geometry

Projective geometry, unlike euclidean geometry, relies on the relationship between objects and their projections. Many foundational concepts including parallelism and transformation significantly differs too. Before we get into some important theorems and topics, lets discuss some of its foundational properties to get the gut feeling for future topics.

First, consider the two, not necessarily similar, triangles and a point \(P\).

[Picture]

Figure 1.1:

From the diagram above, we can think of it as having a light source \(P\) that shines \(\triangle {ABC}\). There, we get our outcome \(\triangle {A'B'C'}\), and 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. With this projection in mind, let’s discuss some fundamental theorems in projective geometry.

1.1 Fundamental Theorems

There are many foundational topics and theorems to discuss. Before we get into more advanced topics, let’s discuss key theorems like Desargues’ Theorem and Pascal’s Theorem.

1.1.1 Desargues’ Theorem

Let’s first get started with Desargues’ Theorem!

Theorem 1.1.1: Desargues’ Theorem

Two triangles are axially perspective if and only if 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 1.1.

[Picture]

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, we normally don’t discuss about that case.

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. With this in mind, let’s try to prove the theorem.

Proof.

The proof of the theorem involves the use of Menelaus’ theorem three times [MATH01-1]. First, let’s 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 can be applied.

[Picture]

Consider \(\triangle {PA'B'}\) with transverse line \(XB\). Using Menelaus’ Theorem, the following equation could be derived. \[ \frac {XA'}{XB'} \cdot \frac {BB'}{BP} \cdot \frac {AP}{AA'} = 1 \]

[Picture]

Similarly, consider \(\triangle {PB'C'}\) and transverse line \(ZC\). \[ \frac {ZB'}{ZC'} \cdot \frac {CC'}{CP} \cdot \frac {BP}{BB'} = 1 \]

[Picture]

With \(\triangle {PC'A'}\) and line \(YC\), \begin{align*} \frac {YA'}{YC'} \cdot \frac {CC'}{CP} \cdot \frac {AP}{AA'} &= 1 \\ \frac {YC'}{YA'} \cdot \frac {CP}{CC'} \cdot \frac {AA'}{AP} &= 1 \end{align*}

Multiplying the equations, \begin{align*} \left ( \frac {XA'}{XB'} \cdot \frac {BB'}{BP} \cdot \frac {AP}{AA'} \right ) \left ( \frac {ZB'}{ZC'} \cdot \frac {CC'}{CP} \cdot \frac {BP}{BB'} \right ) \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*}

is obtained. Continuing, consider the following diagrams.

[Picture]

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 converse must also be proven. Consider the diagram below where the same construction method is used. Note that if \(AA'\), \(BB'\), and \(CC'\) concur is yet to be known.

[Picture]

Notice that \(\triangle {XBB'}\) and \(\triangle {YCC'}\) are in perspective from point \(Z\). Moreover, the first half of the proof manifests 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.

This is it for the proof! Before we move on to the next problem, let’s solve a quick example problem.

Exercise 1.1.2

\(D\) 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.

(Video Solution)

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.