CHAPTER 16

INTRODUCTION TO PROJECTIVE GEOMETRY

16.1 Straightedge Constructions

We saw earlier that a compass alone is as “powerful” as a compass combined with a straightedge. We begin this section by indicating why a straightedge alone is not as powerful as a straightedge and compass or a compass alone. There are only a few admissible operations that can be done with a straightedge by itself.

Admissible Straightedge Operations

A straightedge construction is a finite sequence of the above operations.

We will give informal proofs that certain well-known constructions with straightedge and compass are not possible with a straightedge alone.

One of the standard straightedge and compass constructions is bisecting a given line segment.

Theorem 16.1.1. Using only a straightedge, we cannot construct the midpoint of a given segment.

Proof. The idea behind the proof is that a straightedge construction is projectively invariant. Here we give an intuitive justification of the theorem.

Suppose that there is a finite sequence of the possible straightedge operations that yield the midpoint of a segment AB. In other words, there is a sequence of instructions that, when followed, produces the midpoint of AB. For example, the first few instructions might be:

In the plane , carry out the instructions that yield the midpoint M of the segment AB. Now let ′ be a plane that is not parallel to , as shown in the figure below.

Let O be a point not on or ′ and “project” AB onto ′ from O.

Points in are projected to points in ′. Straight lines in are projected to straight lines in ′. The segment AB in is projected into a segment A′B′ in ′, and each point M of AB projects to a point M′ in A′B′.

Each of the four permissible operations in projects into exactly the same operation in ′. Thus, if a finite sequence of these operations yields the point M of AB, the same finite sequence of instructions carried out in ′ would yield the projected point M′ of A′B′. However, projection from the point O does not preserve midpoints, so it seems that on the one hand the sequence of instructions does yield a midpoint while on the other hand it does not yield a midpoint.

Thus, we see that this construction is possible with a straightedge and compass, but not with a straightedge alone. However, if we have a line parallel to the segment, then we can construct the midpoint of the segment using a straightedge alone, as in the following example.

Example 16.1.2. Given two parallel lines and points A and B on one of them, construct the midpoint of the segment using a straightedge alone.

Solution. Using the straightedge, construct a point C such that the sides AC and BC of ΔABC are cut by the other parallel at E and D, respectively, as in the figure. Let O be the point of intersection of AD and BE, and draw the line through C and O hitting AB at F. Let CF meet m at X. Since l and m are parallel, then ΔCFA ~ ΔCXE, which implies that

so that

that is, that

Also, since ΔCFB ~ ΔCXD, a similar argument shows that

so that

Now, since the three cevians AD, BE, and CF are concurrent at O, and all divisions are internal, from Ceva’s Theorem we have

so that AF = FB and F is the midpoint of AB.

Harmonic Conjugates and Complete Quadrilaterals

What can be constructed using only a straightedge? Problem 6.10 in Section 6.10 illustrates that there are some constructions that can be done.

We will show how to construct the harmonic conjugate of a point using only a straightedge. It is convenient at this point to introduce some terminology.

The figure on the left below consists of four lines a, b, c, and d, no three of which are concurrent, along with the six points formed by the intersections of each pair of lines. Such a figure is called a complete quadrilateral. In the projective plane, some lines may be parallel, in which case some points of intersection may be ideal points. Also, one of the lines may be the ideal line.

The figure on the right illustrates the dual notion, which is a configuration consisting of four points, no three of which are collinear, together with the six lines connecting each pair of points. This configuration is called a complete quadrangle. In the projective plane, one or two of the points may be ideal points, and one of the lines may be the ideal line.

Before we give the construction, we prove a lemma that we alluded to at the beginning of Chapter 15 (Note 2 in Section 15.1).

Lemma 16.1.3. Given four collinear points A, B, C, and D, the points C and D divide AB harmonically if and only if(AB, CD) = −1 or, equivalently, if and only if(CD, AB) = −1.

Proof. Suppose C and D divide AB harmonically. Then

which implies that

However, one of C or D is between A and B while the other is not.

If C is between A and B, we have ; if not, we have = −1. Thus, one of and is positive and the other is negative, which implies that

Conversely, suppose that (AB, CD) = −1. Then

and taking absolute values, this implies that

AC/CB = AD/DB.

Theorem 16.1.4. Given a complete quadrilateral, the two points of intersections of one given diagonal with the other two diagonals divide the vertices of the given diagonal harmonically.

Proof. The following figure illustrates the situation, where the six points of intersection of the sides are A, B, C, D, E, and F, and the three possible diagonals are shown as dashed lines.

Here we wish to show that the points where diagonals ED and AF intersect BC divide BC harmonically; that is, we want to show that (UV, BC) = −1.

Consider the pencils at A and F. We have

Now recall that if (UV, BC) = k, then (UV, CB) = 1/k. Therefore,

so that (UV, BC)² = 1; that is, (UV, BC) is either 1 or −1. Since U and V separate B and C, we must have (UV, BC) = −1.

Example 16.1.5. Given points A, B, and C on a line l, construct the harmonic conjugate D of C using only a straightedge.

Solution. The analysis figure is shown below.

Construction:

Justification:

The fact that D is the harmonic conjugate of C follows from Theorem 16.1.4.

Theorem 16.1.6. Given a circle with unknown center, we cannot construct its center using a straightedge alone.

Proof. The proof is essentially the same as the proof of Theorem 16.1.1.

Using O as the center of projection, project a circle in plane into an ellipse in plane ′. The center C of the circle in plane projects to a point C′ in plane ′, but C′ is not the center of the ellipse in ′.

Let M be the center of the ellipse. Then there is a point C″ in ′ (not shown in the figure) such that M is the midpoint of C′C″. However, a sequence of instructions for the straightedge construction of C in P would “project” into the same sequence of instructions for the straightedge construction of C in ′.

However, with regard to the ellipse in ′, why would the sequence of instructions always yield C′ and never C″? As far as the operations in ′ are concerned, there is no distinction between C′ and C″.

Remark. When it was realized that a straightedge alone could not be used to solve all the construction problems that could be done with a straightedge and compass, the question arose as to what sort of minimal “equipment” was needed in addition to a straightedge. In 1822, Victor Poncelet asserted that all that was needed was a single circle with its center, and in 1833, Jacob Steiner gave a systematic proof of this fact. Such constructions are called Poncelet-Steiner constructions.

Theorem 16.1.7. If A and B are conjugate points on a line that cuts the circle β at C and D, then A and B are divided harmonically by C and D.

Proof. If B is the inverse of A, then this is part of Theorem 14.1.6.

Supposing that B is not the inverse of A, assume that A is outside β. Then B is inside since AB cuts β and B is on the polar of A by assumption. Let A′ be the inverse of A and let α be the circle through A, A′, and B. Recalling that two intersecting circles are orthogonal if one of the circles passes through two distinct points that are inverses with respect to the other circle (see Theorem 13.3.3), this means that α is orthogonal to β.

Since B is on the polar of A, the line A′B is the polar of A, so ∠AA′B is a right angle. From the converse of Thales′ Theorem, AB is a diameter of α, so that β intersects the diameter of α at C and D. By Theorem 14.1.11, β divides the diameter AB of α harmonically; that is, C and D divide A and B harmonically.

The converse to the previous theorem is also true.

Theorem 16.1.8. If a line AB cuts a circle β at C and D, and if C and D divide A and B harmonically, then A and B are conjugate points.

Proof. Assume that A is outside β, and assume that E is a point on the line AB such that A and E are conjugate points. Then C and D divide A and E harmonically by the previous theorem; that is, (AE, CD) = −1.

However, by hypothesis, (AB, CD) = −1, so that

and it follows that B = E; that is, A and B are conjugate points.

The preceding theorems and examples lead to the straightedge construction of the polar of a point with respect to a given circle without its center.

Theorem 16.1.9. Given a circle ω without its center, and given a point P outside ω, it is possible to construct the polar of P using only a straightedge.

Proof. Through P, draw two lines intersecting the circle at B, C and D, E as shown in the figure below.

Let A = BD ∩ CE and F = CD ∩ BE. Then the lines AB, AE, CD, and BE are the sides of a complete quadrilateral with diagonals AF, DE, and BC. From Theorem 16.1.4, AF and DE divide BC harmonically; that is, V is the harmonic conjugate of P, while from Theorem 16.1.8, P and V are conjugate points; that is, V is on the polar of P.

Similarly, H, the intersection of the diagonals AF and DE, is on the polar of P. Thus, VH is the polar of P, and the construction is complete.

Theorem 16.1.10. Let ABCD be a complete quadrangle inscribed in a circle. Let P, Q, and R be the points of intersection of the opposite sides. Then PQR is a self-polar triangle; that is, P is the pole of QR, Q is the pole of PR, and R is the pole of PQ.

Proof. Referring to the figure below, note that the lines RD, RC, AC, and BD form a complete quadrilateral.

Thus, by the proof of the previous theorem, it follows that RQ = p, the polar of P, and similarly, PQ = r. Since Q is on both p and r, then by the reciprocation theorem, P and R are both on q; that is, PR = q.

Another approach to this is given at the end of Section 16.6.

16.2 Perspectivities and Projectivities

Given two planes and ′ and a point O not on either plane, a perspectivity or perspective transformation from onto ′ is a one-one correspondence between points X of and points X′ of ′ such that if X is transformed into X′, then the line XX′ passes through O. The point O is called the center of perspectivity.

The three-dimensional analogue of the projective plane is projective 3-space, which is obtained by appending a “plane at infinity” to Euclidean 3-space. Note that the appended plane is a projective plane.

The same definitions of “perspectivity” and “center of perspectivity” apply to projective 3-space.

Note.

Questions.

Given two lines l and l′ in a plane and a point O not on either line, a perspectivity or perspective transformation from l onto l′ is a one-one correspondence between points X of l and points X′ of l′ such that if X is transformed into X′, then the line XX′ passes through O. The point O is called the center of perspectivity.

If O is an ideal point, as illustrated in the figure on the right above, the result is called a parallel perspectivity. As before, the transformation is one-one and onto.

A finite sequence of perspectivities in a plane is called a projectivity. Note that a perspectivity is always a projectivity, but a projectivity is not necessarily a perspectivity.

Perspectivities and projectivities from one plane to another in 3-space are defined similarly.

We will use lowercase Greek letters to denote perspectivities and projectivities. Two projectivities π₁ and π₂ from l to l′ are equal if π₁(X) = π₂(X) for each X in l. Note that for four perspectivities σ₁ ≠ σ′₁ and σ₂ ≠ σ′₂, it is possible that the projectivities σ₂ σ₁ and σ′₂ σ′₁ are equal.

Effect on Euclidean Properties

In the following table, the properties on the left are preserved by projectivities, and the properties on the right are not preserved by projectivities.

Preserved	Not Preserved
• points	• midpoints
• straight lines	• distance
• collinearity	• angle size
• concurrency	• circularity
• incidence	• order
• triangles
• cross ratios

Note that the table on the previous page indicates that cross ratios are preserved by perspectivities. This fact is an immediate consequence of Theorem 15.1.7, and it plays a crucial role in projective geometry.

The following figure illustrates that order is not preserved.

Projectivities in 2-Space

You may recall from linear algebra that a linear transformation from ⁿ to ⁿ is completely determined by its action on n linearly independent points; that is, if the points

are linearly independent and are mapped respectively to

then this enables us to determine what every other point is mapped to.

There is an analogous situation regarding perspectivities and projectivities that we explore in this section. In particular’, the following questions arise:

Before addressing these questions, we remind ourselves that two projectivities are equal if and only if they have the same effect on every point. The two projectivities could be the composition of different perspectivities, or even different numbers of perspectivities, but that is immaterial to the definition of equality.

Theorem 16.2.1. Given two distinct points A and B on a line l and two distinct points A′ and B′ on a line l′, with l ≠ l′ and none of the four points being l ∩ l′, there is a unique perspectivity that takes A to A′ and B to B′.

Proof. Let O = AA′ ∩ BB′. Then the perspectivity centered at O takes A to A′ and B to B′.

Call this perspectivity π₁, and suppose that π₂ is another perspectivity that takes A to A′ and B to B′. The center of π₂ is AA′ ∩ BB′ = O, and since a perspectivity is completely determined by its center, π₁ and π₂ are identical.

The proof of the corresponding theorem for projectivities is a bit more delicate.

Theorem 16.2.2. Suppose l and l′ are two distinct lines, that A, B, and C are three distinct points on l, and that A′, B′, and C′ are three distinct points on l′. Then there is a unique projectivity from l to l′ that takes A to A′, B to B′, and C to C.

Proof. First we prove existence. We may assume that A ≠ A′ because at least two of A, B, and C differ from l ∩ l′.

Draw a line m through A′ that does not coincide with l′ and that misses A. Pick a point P on AA′ other than A or A′. Using P as the center of perspectivity, map l onto m and denote this perspectivity by σ₁. This takes B to B″ and C to C″.

Let Q = B′B″ ∩ C′C″. Using Q as the center of perspectivity, map m onto l′ and denote this perspectivity by σ₂. Let π = σ₂ σ₁. The projectivity π maps A to A′, B to B′ and C to C′.

Next we prove uniqueness. We have to show that if π₁ and π₂ are projectivities that both take A, B, and C to A′, B′, and C′, respectively, then

for every X in l.

Suppose that for some X we have

Since projectivities preserve cross ratios, we have

so that

which implies that

Theorem 16.2.3. (The Fundamental Theorem of Projective Geometry)

A projectivity in the plane from a line l to a line l′ is completely determined by its action on three distinct points.

Proof. There are two cases to consider: where l and l′ are different and where l = l′. The first case is the previous theorem.

For the case where l = l′, we will show that given three distinct points A, B, and C on l and three more distinct points A′, B′, and C′ also on l, there is a unique projection that takes A to A′, B to B′, and C to C′. To establish the existence of such a projection, draw a line m different than l, and let σ be a perspectivity that maps l onto m. Let A″, B″, and C″ be the images of A, B, and C under σ. By the previous theorem, there is a projectivity π that maps m to l and which takes A″, B″, and C″ to A′, B′, and C′, respectively, The composition π σ is a projectivity from l to l that takes A, B, and C to A′, B′, and C′, respectively. This establishes the existence of a projectivity, and its uniqueness follows as before via cross ratios.

Corollary 16.2.4. A projectivity from a line l to a different line l′ can always be expressed as a sequence of two or fewer perspectivities or a sequence of three or fewer perspectivities if l = l′.

16.3 Line Perspectivities and Line Projectivities

The perspectivities and projectivities described so far are sometimes referred to as being central perspectivities and central projectivities. These two notions can be dualized. Two pencils L and L′ of lines are said to be perspective from a line p if there is a mapping from L to L′ that takes each line x of L to a line x′ of L′ in such a way that p, x, and x′ are concurrent. The line p is called the axis of perspectivity. Such a perspectivity is called a line perspectivity, and a finite sequence of line perspectivities is called a line projectivity.

It is important to realize that any theorem about central projectivities has a dual theorem about line projectivities that we can obtain free of charge. For example, here is the dual of the Fundamental Theorem:

Theorem 16.3.1. A line projectivity from a pencil L to a pencil L′ is completely determined by its action on three distinct lines a, b, and c of L.

When the words projectivity and perspectivity are used without modifiers, they are understood to mean central projectivity and central perspectivity unless the context makes it perfectly obvious that the dual meaning should be used.

16.4 Projective Geometry and Fixed Points

When Is a Projectivity a Perspectivity?

Given a transformation π, any point A for which

is called a fixed point of π.

Note that if π is a transformation from a line l to a line m, then for l ≠ m the only possible fixed point of π is l ∩ m.

Theorem 16.4.1. Suppose l and m are distinct lines. A projectivity π from l to m is a perspectivity if and only if π has a fixed point.

Proof. If π is a perspectivity, then l ∩ m is a fixed point.

On the other hand, suppose that A = l ∩ m is a fixed point of π (the only point that could possibly be a fixed point). Let B and C be two other points on l, and let B′ = π(B) and C′ = π(C).

Now, by Theorem 16.2.1, there is a unique perspectivity σ from l to m that takes B and C on l to B′ and C′ on m, respectively.

However, σ also maps A to A′ = A. Consequently, both π and σ map A, B, and C to A′, B′, and C′, respectively, and so π = σ by the Fundamental Theorem of Projective Geometry.

Theorem 16.4.2. A projectivity π from a line to itself is a perspectivity if and only if it has at least three fixed points.

We leave the proof as an exercise.

Alternate Characterizations

Theorem 16.4.3. Suppose that

are two perspectivities with different centers. Then σ π is a perspectivity if and only if k, l, and m are concurrent.

Proof. Let A and A′ be the centers of perspectivities for π and σ.

There are two things to prove:

We will prove (1) and leave (2) as an exercise.

Assuming that σ π is a perspectivity from k to m, let P be the center of the perspectivity.

Let B and C be distinct points on k, and let

and

Then P = BB′ ∩ CC′.

Now draw the line AA′ and let D, D″, and D′ be defined as follows:

Then π maps D to D″ and σ maps D″ to D′.

Consequently, σ π maps D to D′, and since σ ∩ π is really a perspectivity with center P, the line DD′ must pass through P. Also, since , this means that passes through P.

Therefore, ΔABC and ΔA′B′C′ are perspective from P; that is, they are copolar. Hence, the triangles are perspective from a line; that is, they are coaxial by Desargues′ Two Triangle Theorem. However,

and so the axial line is B″C″, namely, l. Since BC = k and B′C′ = m, the point k ∩ m is on l, showing that k, l, and m are concurrent.

16.5 Projecting a Line to Infinity

Straight lines and incidence properties are preserved by projections, and this is sometimes advantageous. In the figure below, the quadrilateral ABCD lies in a plane . Another plane ′ intersects in a line parallel to the diagonal BD of the quadrilateral. The point O is the center of a perspectivity from to ′, and O is positioned so that the plane defined by the points O, B, and D is parallel to ′. With this perspectivity, points B and D are projected to ideal points B′ and D′ since the lines OB and OD are parallel to ′. Points A and C are projected to ordinary points A′ and C′. In this particular case, C′ = C, since the point C happens to be on the line ∩ ′.

In the plane ′, the line B′D′ is the ideal line, that is, the line at infinity. Consequently, the process above is called sending the line BD to infinity.

The projected lines A′B′ and C′B′ are necessarily parallel, as are the lines A′D′ and C′D′. This is all that we need to know in order to depict the figure with the line sent to infinity. In figure (a) below we have quadrilateral ABCD, with its projected image A′B′C′D′ shown in figure (b), and there is really no need to draw the planes and ′.

We illustrate the use of this technique to prove the following version of Pappus′ Theorem.

Theorem 16.5.1. If the three lines APP′, UQQ′, and URR′ meet two lines OX and OX′ at P, Q, R and P′, Q′, R′, respectively, then the points

are collinear.

Proof. Send the line OU to infinity. In the diagram below, we use the same letter before and after the projection.

The lines PP′, QQ′, and RB′ are parallel, as are PR and P′R′. Then PQ′ ∩ P′Q is the intersection of the diagonals of the parallelogram PQQ′P′. Similarly, PR′ ∩ P′R and QR′ ∩ Q′R are the intersections of the diagonals of parallelograms, so all three points lie on a line midway between PR and P′R′.

As another example, we use the same technique to prove part of Desargues′ Theorem.

Example 16.5.2. Show that copolar triangles are coaxial.

Solution. In the figure below, triangles ABC and A′B′C′ are perspective from the point O. We want to show that the intersections of the corresponding sides, namely, P, Q, and R, are collinear.

A solution may be obtained by sending the line QR to infinity and then showing that the projection also sends point P to the ideal line. The reverse projection then maps the ideal line back to the original line QR, and since it preserves incidence, the point P must lie on the line QR.

The projection of the line QR to infinity is shown below, As before, we use the same letters before and after the projection.

For triangles OBC and OB′C′, since BC and B′C′ are parallel, we have

For triangles OAC and OA′C′, since AC and A′C′ are parallel, we have

Consequently,

which implies that AB is parallel to A′B′ and, thus, the projected point

is on the ideal line.

16.6 The Apollonian Definition of a Conic

Originally, the ancient Greeks defined conic sections as cross sections of particular types of right circular cones. Apollonius recognized that all of the conic sections can be obtained from any given circular cone. He defined a circular cone as follows:

The line through K that passes through the center of ω is called the axis of the cone. If the axis of the cone is perpendicular to the plane of the circle, the cone is called a right circular cone, otherwise the cone is called an oblique circular cone. Note that the cone has two components, so they are often called two-napped cones.

Apollonius defined a (proper) conic section as a curve that is formed by intersecting a plane with a circular cone in such a way that the plane does not contain the vertex K of the cone. The figure above shows what happens as the tilt of the plane is increased. When the plane is tilted so that it intersects only one nappe of the cone and is not parallel to any of the generating lines of the cone, the result is an ellipse. If the plane is parallel to one of the lines generating the cone, but it still only intersects one nappe, the result is a parabola. When it intersects both nappes of the cone, the result is a hyperbola.

It is not too difficult to derive the Cartesian equation of the conies from Apollonius′ description. Here is how it is done for the ellipse. In the figure on the following page, the plane ξ, and any plane parallel to it, cuts the oblique circular cone in a circle. The plane π is another plane that cuts the cone, forming the curve α The planes ξ and π meet in a straight line m.

In the plane ξ, the line AB is a diameter of the circle and is perpendicular to m and C = AB ∩ m. The plane of the triangle ABK cuts the plane π in the line EDC. To obtain a Cartesian equation for α, let P be an arbitrary point on α, and through P pass a plane ϕ (not shown in the diagram) parallel to ξ cutting the cone in a circle ω, as shown in figure (a) above.

The intersection of ϕ and π is a line parallel to m cutting ω at P and Q. The plane ϕ cuts AK at F and BK at G, so that FG is a diameter of the circle ω. The lines ED, FG, and PQ are concurrent at H.

Let E be the point (0, 0) in π, let x be the distance EH, and let y be the distance PH so that P is the point (x, y). Since PQ ⊥ FG, by the power of the point H with respect to ω (see figure (b) above), we have PH² = FH · HG. This gives us

(1)

From similar triangles EFH and EAC, we have FH/AC = EH/EC, so that

(2)

From similar triangles HGD and CBD, we have HG/CB = HD/CD. Since HD = ED − x, we get

(3)

Letting ED = 2a, and substituting (2) and (3) into (1), we obtain

Denoting the positive quantity (AC · CB)/(EC · CD) by k², we have

and so

This shows how the Cartesian equation for the ellipse arises from the Apollonian definition of the conic.

Poles and Polars of Conies

The Apollonian definition of conies can be described by saying that every conic can be interpreted as being the image of a circle under a central perspectivity, Thus, we can state without proof:

Theorem 16.6.1. Every theorem about the incidence properties of straight lines and circles remains true when the word “circle” is replaced by the word “conic.”

For example, both Pascal’s Mystic Hexagon Theorem and Brianchon’s Theorem are true for parabolas, ellipses, and hyperbolas.

One of the important consequences of Theorem 16.6.1 is that in order to prove something about the incidence properties of straight lines and conies, we can reduce it to proving the same assertion about circles. There is a caveat, however, in that the properties must be described purely in terms of incidence properties.

For example, the notions of pole and polar for a circle are useful concepts that were not initially described in terms of incidence properties. If P is a point outside a circle ω, the polar of P can be described using incidence properties: let A and B be the points where the tangents from P meet the circle. Then is the polar of P. If P is a point inside the circle, a description of the polar in terms of the incidence is not immediately evident.

Theorem 16.6.2. Let P be a point not on the circle ω, and let l and m be lines through P, with l ∩ ω = {A, B} and m ∩ ω = {C, D}. Let a, b, c, and d be the tangent lines through A, B, C, and D. Then the line through a ∩ b and c ∩ d is the polar of P.

Proof. Let L = a ∩ b and M = c ∩ d. Then L is the pole of l, and M is the pole of m. Since P is on l and m, the Reciprocation Theorem implies that L and M are on p.

We indicated earlier that the polar of a point P on a conic is the line p that is tangent to the conic at P. The lemma on the following page allows us to describe the polar of a point P not on the conic entirely in terms of incidence properties.

Definition. Suppose ω is a conic. If P is not on the conic, then the polar of P is the line p described in the preceding theorem with the word “circle” replaced by the word “conic”.

Theorems 16.1.8 and 16.1.9 illustrate how to construct the polar using only a straightedge and without necessarily drawing tangents. The next few lemmas provide an alternate approach that illustrates the connection with Pascal’s Mystic Hexagon Theorem. In the lemmas, we follow the convention that lowercase letters refer to the polars of points designated by the corresponding uppercase letters.

Lemma 16.6.3. Suppose that A, E, B, and D are four distinct points on a circle ω. Let P = AE ∩ DB and Q = AB ∩ DE. Let S = e ∩ b and T = a ∩ d Then P, Q, S, and T are collinear.

Proof. Introduce points F and C on the circle, as shown in the figure below, so that ABCDEF is an inscribed hexagon. By Pascal’s Mystic Hexagon Theorem, the opposite edges meet in three collinear points Q, M, and N, as depicted in the figure.

Keep the points A, E, B and D fixed, and let F → E and C → B along the circle. As this happens, M and N move but Q, M, and N remain collinear. The limiting points of M and N are S and P, respectively, so Q, S, and P are collinear.

Now apply a similar procedure to A and D, as shown on the following page. Here, the points M, Q, and N are collinear.

Letting F → A and C → D along the circle, the points M and N converge to P and T, respectively, and so P, Q, and T are collinear.

Consequently P, Q, S, and T are collinear.

The following is proven in a similar way:

Lemma 16.6.4. Suppose that A, E, B, and D are distinct points on a circle, as in the previous lemma. Let R = AD ∩ EB and let Q = AB ∩ DE, as in the previous lemma, and let U = a ∩ e and V = d ∩ b. Then R, Q, U, and V are collinear.

Theorem 16.6.5. Suppose that A, E, B, and D are distinct points on a circle. Let P = AE ∩ BD, Q = AB ∩ DE, and R = AD ∩ BE. Then RQ is the polar of P, PQ is the polar of R, and PR is the polar of Q.

Proof. In the proof we continue to employ the convention that lowercase letters refer to the polars of points designated by the corresponding uppercase letters.

Introduce the polars a, b, c, and d, and let U = a ∩ e, S = e ∩ b, V = b ∩ d, and T = a ∩d.

By the previous lemmas, the points P, Q, S, and T are collinear, as are R, U, Q, and V. By Theorem 16.6.2, UV is the polar of P, which implies that RQ is the polar of P, as claimed. Similarly, ST is the polar of R, which implies that PQ is the polar of R.

Since Q is on r, R must be on q by the Reciprocation Theorem. Similarly, since Q is on p, P must be on q. That is, R and P are on q, so RP = q.

16.7 Problems