CHAPTER 15

CROSS RATIOS

15.1 Cross Ratios

Directed Distances

Recall that we denote directed distances, or signed distances, with a bar over the distance, as in , and that

Also recall the following facts:

Directed Distances and Harmonic Conjugates

Recall that given points A, B, C, and D, then B and D are harmonic conjugates with respect to A and C if and only if

as in the figure below.

Here, we are using unsigned distances; for signed distances, B and D are harmonic conjugates with respect to A and C if and only if

Properties of Cross Ratios

Given a range of four points, A, B, C, and D, we define the quantity (AB, CD) by

and call it the cross ratio of the points A, B, C, and D, taken in that order.

Note.

Example 15.1.1. Find (AB, CD), (AC, BD), and (BA, DC) where A, B, C, and D are collinear with coordinates along the line given by 0, 1, 2, and 3, respectively.

Solution. Working directly from the definition of cross ratios, we get

Similarly,

and

 

Theorem 15.1.2. If (AB, CD) = k, then:

Proof. (1) and (2) follow directly from the definition of the cross ratio. To prove (3), we will use the fact that for three collinear points X, Y, and Z, the directed distances are related by = + , whether Y is between X and Z or not.

Interchanging the middle pair, we have

Interchanging the outer pair, from (1) we have

and interchanging the middle pair on the right-hand side, we have

and again by (1), we have

Remark. Any permutation of the letters A, B, C, and D can be obtained by successively interchanging pairs using (1), (2), or (3) of Theorem 15.1.2.

Example 15.1.3. Given (AB, CD) = k, find (DA, CB).

Solution. From (3), we have

while from (2), we have

Alternatively, we have

which implies that

which in turn implies that

Ideal Points

Suppose that one of the four points A, B, C, or D is an ideal point I. We use the convention that

For example, if B = I, then

Now suppose we interchange the two middle symbols in (AI, CD). Then by direct computation we get

However,

which shows that the theorem on permutation of symbols remains true if one of the points is an ideal point.

There are 24 different arrangements of the symbols A, B, C, and D, giving rise to 24 different cross ratios: (AB, CD), (BA, CD), (BC, AD), and so on. However, there are only six different values for the 24 cross ratios, namely,

To see why this is so, note that all permutations of the symbols A, B, C, and D can be obtained via a sequence of interchanges of the types described in Theorem 15.1.2 and that only the operations of the second and third types produce different values. Thus, if the cross ratio (WX, YZ) has the value k, then the only new values we can obtain by operation of the second or third type are 1/k and 1 − k. The value 1/k in turn can produce values of k or 1 − 1/k = (k − 1)/k. If we were to continue in this manner, we would see that only the six different values listed above can be obtained.

Theorem 15.1.4. If (AB, CD) = (AB, CX). then D and X are the same point.

Proof. We have

so that

which implies that

From the remarks at the beginning of the chapter on the ratios of directed distances, this implies that D = X.

 

Corollary 15.1.5. If (AB, CD) = (AB, XD), then C = X.

Proof. We have

so that

and from the previous theorem, C = X.

Cross Ratio of a Pencil of Lines

Let k, l, m, and n be a pencil of lines concurrent at a point P. This section will provide a definition of the cross ratio, denoted (kl, mn), of the pencil of lines.

Point of Concurrency is an Ordinary Point

Suppose k, l, m, and n are concurrent at an ordinary point P, and let A, B, C, and D be points other than P on k, l, m, and n, respectively, as in the figure below.

We define P(AB, CD) as follows:

where the directed angle is the angle from the ray to the ray and whose magnitude is between 0° and 180°.

Note that if A and A′ are points on k on the opposite sides of P, then the directed angles and are different, as are the directed angles and . In fact,

Since sin(x − 180) = −sin x, it follows that if A and A′ are on opposite sides of P on the line k, then

or, in other words,

Thus, we have the following result:

Theorem 15.1.6. For a pencil of lines k, l, m, and n that are concurrent at the ordinary point P, the definition of P(AB, CD) is independent of the choice of the points A, B, C, and D, as long as none of them are the point P.

This allows us to make the following definition:

Definition. We define the cross ratio of a pencil of lines concurrent at an ordinary point P as

where A, B, C, and D are points on k, l, m, and n other than P.

Point of Concurrency Is an Ideal Point

When the point of concurrency is an ideal point, the lines k, l, m, and n are parallel.

In this case, let t be any line intersecting k, l, m, and n at A, B, C, and D, respectively, and define (kl, mn) to be (AB, CD).

To check that this definition is independent of the choice of the line t, let t′ be another line intersecting k, l, m, and n at A′, B′, C′, and D′, There are two cases to consider:

Case (i). t and t′ are parallel.

In this case, obviously AC = A′C′, CB = C′B′, etc., since they are opposite sides of a parallelogram, so that

Case (ii). t and t′ meet at an ordinary point P.

In this case, by similar triangles,

with similar results for the other ratios. From this it follows that

Theorem 15.1.7. Suppose that k, l, m, and n form a pencil of lines concurrent at the ordinary point P. If a transversal cuts the lines k, l, m, and n at the points A, B, C, and D, respectively, then

Proof. In order to prove the theorem, we will show two things:

Proof of (1).

To check that (1) is true, note that given a directed line t and a point P not on t, then either

for all pairs of points X and Y on t or else

for all pairs of points X and Y on t.

For points A, B, C, and D on t, the value of (AB, CD) is independent of the direction of t (see statement (3) in the note following the definition of (AB, CD) at the beginning of this chapter). In particular, we can choose the direction of t so that for all pairs of points X and Y,

Thus,

Proof of (2).

The proof uses the fact that it is possible to compute the area of a triangle in two different ways:

In the figure above,

where (b) follows from the fact that QY = PY sin(∠XPY).

Expanding |(AB, CD)|, we get

Calculating the areas using (b) we get

Substituting these values into the expression for |(AB, CD)| above, we get

This completes the proof of the theorem.

15.2 Applications of Cross Ratios

Four Useful Lemmas

The following lemmas connect cross ratios with concurrency and collinearity.

Lemma 15.2.1. In the figure below, P is an ordinary point and P, A, and A′ are collinear; P, B, and B′ are collinear; and P, C, and C′ are collinear. Transversals cut the lines PA, PB, and PC at A, B, and C, respectively. The point D is on AC, and the point D′ is on A′C′.

With this configuration, if (AB., CD) = (A′B′, C′D′), then P is on DD′.

Proof. Suppose the line PD intersects A′D′ at some point E′ (E′ is not shown in the diagram). Then from Theorem 15.1.6, we have

However,

so that

which implies that E′ = D′, and P is on DD′.

Lemma 15.2.2. In the figure below, two lines intersect at the point A. The points B, C, and D are on one of the lines, while the points B′, C, and D′ are on the other line.

With this configuration, if(AB, CD) = (AB′, C′D′), then BB′, CC, and DD′ are concurrent.

Proof. Let P = BB′ ∩ CC′ and let E′ = PD ∩ AC′, as shown below.

It suffices to show that E′ = D′.

We have

which implies that

and since (AB, CD) = (AB′, C′D′), it follows that

Therefore, E′ = D′.

 

Lemma 15.2.3. In the figure below, A, B, and C are points on a line m. P and Q are points not on m. D is a point other than A, B, C, P, or Q.

With this configuration, if P(AB, CD) = Q(AB, CD), then D is on m.

Proof. If D is not on m, let PD intersect m at X and let QD intersect m at Y, as shown below.

We have

and

and since P(AB, CD) = Q(AB, CD), we must have (AB, CX) = (AB, CY).

However, this can only happen if PD ∩ QD = X = Y = D; that is, D is on m.

Lemma 15.2.4. In the figure below, A, B, C, and D are points other than P and Q, and the point A is on PQ.

With this configuration, if

then B, C, and D are collinear.

Proof. Let A′ be the point PQ ∩ BC and let m be the line BC. Then

which implies that

which implies that D is on m, by the previous lemma.

Theorems of Desargues, Pascal, and Pappus

Desargues′ Two Triangle Theorem (Theorem 4.4.3) was proven in Part I using Ceva’s Theorem and Menelaus′ Theorem. We restate it here and prove it using cross ratios.

Theorem 15.2.5. Copolar triangles are coaxial and conversely.

Proof. Let the copolar triangles be ABC and A′B′C′. Then, as in the figure on the following page,

Let P, Q, and R be the points of intersection of the corresponding sides of the triangle. In order to show that P, Q, and R are collinear, we will show that R is on the line PQ.

Let X, X′, and Y be as shown; that is, let X = OC′ ∩ AB, X′ = OC′ ∩ A′B′, and Y = OC′ ∩ PQ.

There are three pencils of lines: one concurrent at C, one concurrent at O, and one concurrent at C′.

Since BA is a transversal for the pencil at C, we have

and since B′A′ is a transversal for the pencil at O, we have

and replacing B′ by P, X′ by Y, and A′ by Q in the last expression, we have

Therefore, C(PY, QR) = C′(PY, QR), and by Lemma 15.2.4, P, Q, and R are collinear.

Pascal’s Mystic Hexagon Theorem from Part I (Theorem 4.4.4) says the following:

Theorem 15.2.6. If a hexagon is inscribed in a circle, the points of intersection of the opposite sides are collinear.

There are many possible configurations, three of which are shown in the following figure.

In order to prove the theorem using cross ratios we first need the following lemma:

Lemma 15.2.7. If A, B, C, D, and P, Q are distinct points on a circle, then P(AB, CD) = Q(AB, CD).

Proof. There are two cases to consider, as illustrated in the following figures.

Case (1). P and Q are not separated by any of the points A, B, C, or D.

In this case, Thales’ Theorem implies that

Thus,

Case (2). P and Q are separated by some of the points A, B, C, or D.

The proof here is similar to that for Case (1), but now we have

The positive signs in the second and third equations arise since the signed angles are in opposite directions. From the second equation, we get

Similarly, from the third equation, we get

The lemma now follows from the definitions of P(AB, CD) and Q(AB, CD) as in Case (1).

 

For convenience, we restate Pascal’s Theorem.

Theorem 15.2.8. If ABCDEF is a hexagon inscribed in a circle, then the points of intersection of the opposite sides are collinear.

Proof. The proof works for any configuration, and for clarity we use the one in the figure below.

Let X = AB ∩ CD and Y = BC ∩ AF.

Consider the pencils at D and F. Since P is on DE and X is on DC, then

and from Lemma 15.2.7, we have

Since Y is on FA and Q is on FE, then

and, therefore,

Note that we have the following configuration:

Since (AP, XB) = (YQ, CB), then from Lemma 15.2.2 it follows that AY, PQ, and XC are concurrent. However, AY ∩ XC = R, so PQ passes through R.

 

As a final example, we prove Pappus’ Theorem (Theorem 4.4.5) using cross ratios:

Theorem 15.2.9. Given points A, B, and C on a line l and points A′, B′, and C′ on a line l′, then the points of intersection

are collinear.

Proof. Introduce points X = AC′ ∩ A′B and Y = CA′ ∩ C′B, as shown in the figure below.

Using the pencil through A, replace B by O, X by C′, and P by B′. Then

and since these are transversals for the pencil through A, we have

Using the pencil through C, replace O by B, B′ by R, and A′ by Y. Then

and since these are transversals for the pencil through C, we have

Thus,

It follows from Lemma 15.2.2 that XC′, PR, and A′Y are concurrent, and since XC′ ∩ A′Y = Q, then Q is on PR; that is, P, Q, and R are collinear.

15.3 Problems