CHAPTER 4

THEOREMS OF CEVA AND MENELAUS

4.1 Directed Distances, Directed Ratios

This chapter provides a more algebraic approach to concurrency and collinearity through the theorems of Ceva and Menelaus. The theorems are best understood using the notions of directed distances and directed ratios. We repeat the definition given in the previous chapter.

Let l be a line and assign a direction to the line For two points A and B on the line, with A ≠ B, let AB be the distance between A and B. The directed distance or signed distance from A to B, denoted , is defined as follows:

The directed distances and for the given direction along l are shown in the figure below, where AB = 2.

Properties of Directed Distance

Sometimes property (2) is stated as . Note that properties (2) and (3) do not hold for unsigned distances.

Ratios

In the theorems of Ceva and Menelaus, given three points A, B, and C on a line, the following directed ratio occurs frequently:

Note that if C is between A and B, then

while if C is external to the segment AB, then

4.2 The Theorems

The two theorems are considered as companions of each other, although their discoveries were separated by many centuries. Menelaus proved his theorem around the year 100 CE. It languished in obscurity until 1678, when it was uncovered by Giovanni Ceva, who published it along with the theorem that bears his name. The two theorems are strikingly similar, and it is surprising that there was such a time span between the two discoveries.

Theorem 4.2.1. (Ceva’s Theorem)

Let AD, BE, and CF be lines from the vertices A, B, and C of a triangle to nonvertex points D, E, and F of the opposite sides. The lines are either concurrent or parallel if and only if

Given a triangle, any line that passes through a vertex of the triangle and also through a nonvertex point of the opposite side is a called a cevian line or cevian of the triangle.² Medians, angle bisectors, and altitudes are cevians, but the right bisector of a side of a nonisosceles triangle is not. Also, a line through a vertex parallel to the opposite side of a triangle is not a cevian, although we will later incorporate this when we discuss the extended Euclidean plane.

Given a triangle, a nonvertex point in a side or in an extended side of the triangle is called a menelaus point. A line that passes through each of the three edges of a triangle, but not through any of the vertices, is called a transversal line or a transversal. Menelaus’ Theorem tells us when three menelaus points lie on a transversal.

Theorem 4.2.2. (Menelaus’ Theorem)

Let D, E, and F be menelaus points on the (extended) sides BC, CA, and AB of a triangle ABC. The points are collinear if and only if

To use the two theorems, simply assign a direction to each side of the triangle, and proceed around the triangle in either a clockwise or counterclockwise direction. The product of the ratios in the theorems is sometimes called the cevian product.

Note that the cevian product

involves the directed ratios

These ratios are meaningful because points F, D, and E are in the (perhaps extended) sides AB, BC, and CA of the triangle, respectively.

Note also that changing the direction assigned to the sides of the triangle does not affect the sign of the directed ratio. It should also be noted that if you compute the cevian product by proceeding clockwise rather than counterclockwise, the two cevian products will be reciprocals of each other. This does not alter the theorems because +1 and −1 are the only two real numbers that are their own reciprocals!

The original versions of the theorems used undirected distances, and, although the theorems are often stated that way, the modern versions are easier to use.

Ceva’s Theorem will be the object of study in the next section, and Menelaus’ Theorem will be discussed in the section following that.

4.3 Applications of Ceva’s Theorem

The theorems of Ceva and Menelaus involve directed ratios of the form

where X is a point on the line AB other than A or B. Consequently, either X divides AB internally (meaning that X is between A and B) or else X divides AB externally (X is not between A and B), It is useful to recall that in these cases,

As well, it is convenient to define directed ratios for segments that belong to different but parallel lines. Thus,

This is illustrated in the figure on the following page.

Before proving Ceva’s Theorem, we will illustrate how it can be used to show that the most familar cevians are concurrent. Some of the results use the so-called Crossbar Theorem, which we state without proof:

Theorem 4.3.1. (Crossbar Theorem)

If P is an interior point of triangle ABC, then the ray meets side B at some point between B and C.

A frequently used consequence, that we again state without proof, is:

Theorem 4.3.2. Two cevians that pass through the interior of a triangle are not parallel.

Here is how Ceva’s Theorem can be used to prove that medians, angle bisectors, and altitudes are concurrent.

Example 4.3.3. The medians of a triangle are concurrent.

Solution. We have

By Ceva’s Theorem, the medians are either concurrent or parallel. Since they cannot be parallel by Theorem 4.3.2, they must be concurrent.

Example 4.3.4. The internal bisectors of the angles of a triangle are concurrent.

Solution.

Recall from the Angle Bisector Theorem that if the interior angle bisector of ∠A meets side BC at D, then BD/DC = AB/AC. Therefore,

By Theorem 4.3.2, the cevians in this case cannot be parallel, so by Ceva’s Theorem, they are concurrent.

Example 4.3.5. The altitudes of a triangle are concurrent.

Solution.

For a right triangle, the three altitudes are concurrent at the vertex of the right angle. For an acute-angled triangle, the three altitudes divide the sides internally, and for an obtuse-angled triangle, the altitudes divide exactly two of the sides externally. (A proof of this may be obtained from the External Angle Inequality and is left as an exercise.)

Therefore, for a triangle ABC with altitudes AD, BE, and CF, we have

since either all ratios are internal or else exactly two are external.

Using similar triangles to relate the ratios to the length of altitudes,

Hence,

and Ceva’s Theorem shows that the altitudes are concurrent.

The next example illustrates how Ceva’s Theorem can be used to compute a ratio or a distance rather than to conclude that certain cevians are concurrent.

Example 4.3.6. In the figure, AB = 4, BC = 5, and AC = 6. AD is an angle bisector, and BE is a median. Find the length of AF.

Solution. By Ceva’s Theorem,

By the Angle Bisector Theorem, BD/DC = 4/6, and since BE is a median, CE/EA = 1, so that

from which we get AF = 12/5.

4.4 Applications of Menelaus’ Theorem

Angle Bisectors

There are three internal bisectors and three external bisectors of the angles of a triangle. If we take one bisector, either internal or external, from each vertex, then there are four possible combinations:

For (i) and (ii), the bisectors are concurrent, and so it is probably no surprise to the reader that there is something significant about (iii) and (iv).

Example 4.4.1. The internal bisectors of two angles of a triangle and the external bisector of the third meet the opposite sides in three collinear points.

Solution. Let us consider the internal bisectors of ∠A and ∠B and the external bisector of ∠C. Using the Internal and External Bisector Theorems,

and so by Menelaus’ Theorem, the points are collinear.

Simson’s Theorem

The following proof of Simson’s Theorem uses Menelaus’ Theorem.

Example 4.4.2. Given ΔABC and a point P on its circumcircle, the perpendiculars dropped from P meet the sides of the triangle in three collinear points.

Solution. Let D, E, and F be the feet of the perpendiculars on BC, CA, and AB, respectively. Since D, E, and F are three menelaus points of ΔABC, it is enough to show that the cevian product is − 1.

Introducing PA, PB, and PC, Thales’ Theorem reveals that ∠PAF ≡ ∠PCD. Since both ΔPAF and ΔPCD are right triangles, they are similar. In fact, we find that

with the last similarity following from Theorem 1.3.11. It follows that

and the result is that

To check that the sign is actually negative, we have to show that an odd number of the points D, E, and F divide the sides externally. This may be accomplished by constructing a hexagon AQBRCS whose sides are perpendicular to the sides of ΔABC, as in figure (a) on the following page.

Then, the vertices Q, R, and S are on the circumcircle of ΔABC, since the quadrilaterals AQBC, BRCA, and CSAB are cyclic.

Referring to figure (a), it is not difficult to see that if P is strictly between A and Q on the small arc AQ, then the perpendicular PE from P to AC divides AC externally. (The reason is that in order to divide AC internally, the point P would have to be between the lines AQ and CR.) On the other hand, the perpendiculars from P to AB and BC divide those sides internally.

Similar reasoning applies when P is one of the other six arcs of the circumcircle and also to the case where the hexagon is nonconvex, as in figure (b) above.

Desargues’ Two Triangle Theorem

In this section we will prove Desargues’ Two Triangle Theorem, but first the definitions of copolar and coaxial triangles.

Two triangles ΔABC and ΔA′B′C′ are said to be copolar from a point O if and only if the joins of corresponding vertices are concurrent at the point O; that is, if and only if the lines AA′, BB′, and CC′ are concurrent at O, as in the figure above. The point O is called the pole.

Two triangles ΔABC and ΔA′B′C′ are coaxial from a line l if and only if the points of intersection of corresponding sides

are collinear, as in the figure above.

Theorem 4.4.3. (Desargues’ Two Triangle Theorem)

Two triangles in the plane are copolar if and only if they are coaxial, as in the figure below.

The line l is called the Desargues line.

Proof. Copolar implies Coaxial.

Given that ΔABC and ΔA′B′C′ are copolar from O, we need to show that

are collinear, and from Menelaus’ Theorem we only have to show that

We will use a decomposition type argument. We decompose ΔBOC into three triangles ΔAOB, ΔAOC, and ΔABC and apply Menelaus’ Theorem to three transversals as follows:

Multiplying these three expressions together, we have

where

Thus, X = 1, so that

and P, Q, and R are collinear.

Coaxial implies Copolar.

Given a pair of coaxial triangles ΔABC and ΔA′B′C′, as in the figure on the following page, we want to show that they are copolar; that is, we want to show that AA′, BB′, and CC′ are concurrent.

Let O be the intersection of BB′ and CC′. Then we have to show that AA′ also goes through O.

Observe that ΔQBB′ and ΔRCC′ are copolar from P. From the first half of the proof, we know that ΔQBB′ and ΔRCC′ are also coaxial, so that

are collinear, and so AA′ passes through O. Therefore, ΔABC and ΔA′B′C′ are copolar from O.

Pascal’s Mystic Hexagon Theorem

In this section we will prove Pascal’s Mystic Hexagon Theorem, but first we give a convention about labeling the sides of a hexagon, simple or nonsimple.

Given a hexagon ABCDEF, simple or nonsimple, inscribed in a circle, we label the sides with the positive integers so that

as in the figure on the following page.

We say that sides 1 and 4. sides 2 and 5, and sides 3 and 6 are opposite sides, whether the hexagon is simple or nonsimple.

Theorem 4.4.4. (Pascal’s Mystic Hexagon Theorem)

Given a hexagon (simple or nonsimple) inscribed in a circle, the points of intersection of opposite sides are collinear and form the Pascal line.

Proof. Given an inscribed hexagon as shown, we want to show that P, Q, and R are collinear.

Create ΔXYZ by taking every second side of the hexagon so that

Note that

so that P, Q, and R are menelaus points of ΔXYZ.

In order to show that P, Q, and R are collinear, by Menelaus’ Theorem we need only show that

Applying Menelaus’ Theorem to the following labeled transversals of ΔXYZ, we have

and multiplying these together we get

Now, the power of the point X with respect to the circle is

the power of the point Y with respect to the circle is

while the power of the point Z with respect to the circle is

and therefore,

so that

and the points P, Q, and R are collinear.

Pappus’ Theorem

In this section, we will prove Pappus’ Theorem. Although it belongs to the realm of projective geometry, we can give an elementary proof using Euclidean geometry via Menelaus’ Theorem.

Theorem 4.4.5. (Pappus’ Theorem)

Given points A, B, and C on a line l and points A′, B′, and C′ on a line l′, the points of intersection

are collinear, as in the figure.

Proof. Extending the sides of the triangle formed by the sides A′B, B′C, and C′A, we have the figure on the following page.

We will apply Menelaus’ Theorem to each of the five transversals of the triangle ΔUVW

in turn.

For the transversal APB′, we have

by Menelaus’ Theorem.

For the transversal CQA′, we have

by Menelaus’ Theorem.

For the transversal BRC′, we have

by Menelaus’ Theorem.

For the transversal ABC, we have

by Menelaus’ Theorem.

For the transversal A′B′C′, we have

by Menelaus’ Theorem.

Multiplying the first three expressions together, we get

where

Rearranging the terms in X, we get

from the last two expressions, and therefore,

so that P, Q, and R are collinear by Menelaus’ Theorem.

Numerical Applications

In addition to using Menelaus’ Theorem to determine when menelaus points are collinear, we can also use it to calculate distances when three points are known to be collinear.

Such problems often present us with several different ways of viewing the same diagram.

In the figure on the right, there are several “menelaus patterns”:

• AB is a transversal for ΔPEC.
• AC is a transversal for ΔPDB.
• BE is a transversal for ΔADC.
• CD is a transversal for ΔABE.

In solving a problem, it is sometimes a matter of trying various combinations until we find one that works.

Example 4.4.6. BD is a median of ΔABC, and E is the midpoint of BD. Line CE extended meets AB at F. If AB = 6, find the Length of FB.

Solution. We wish to find a menelaus pattern whose triangle has two of its edges divided into known ratios. Referring to the figure, note that if we consider CF as a transversal for ΔABD, then CF divides the sides BD and AD into known ratios. So, let FB = x and apply Menelaus’ Theorem to ΔABD with transversal CF:

which implies that

which in turn implies that

Thus, AF = 2x, so AB = 3x, and it follows that x = AB/3 = 2.

4.5 Proofs of the Theorems

Both Ceva’s and Menelaus’ Theorems are of the “if and only if” variety, and so in each case there are two things to prove.

Proof of Ceva’s Theorem

We will first prove the necessity; that is, we will prove the following:

Theorem 4.5.1. (Ceva’s Theorem: Necessity)

Let AD, BE, and CF be cevians of triangle ABC, with D on BC, E on CA, and F on AB. If AD, BE, and CF are concurrent or parallel, then

Proof. We will prove the theorem for the case where AD, BE, and CF are concurrent. The case where the cevians are parallel is left as an exercise. Suppose the cevians AD, BE, and CF are concurrent at a point P. Let l be a line through A parallel to BC, let S be the point where BE meets l, and let R be the point where CF meets l.

Then

with the last equality arising from the fact that

because ΔBPD ~ ΔSPA and ΔDPC ~ ΔAPR. It follows that

which completes the proof of necessity.

Next, we wish to prove the sufficiency, that is:

Theorem 4.5.2. (Ceva’s Theorem: Sufficiency)

Let AD, BE, and CF be cevians of triangle ABC, with D on BC, E on CA, and F on AB. If

then AD, BE, and CF are either concurrent or parallel.

Proof. There are two cases to consider: either (i) BE and CF are parallel or else (ii) BE and CF meet at a single point P.

Case (i). Let l be a line through A parallel to BE. Since BE is not parallel to BC (by the definition of a cevian), it follows that l is not parallel to BC, and so l must meet the line BC at some point G. Then, the cevians AG, BE, and CF are parallel, so by the “necessary” part of the theorem we must have

By hypothesis, we also have

so that

Thus,

and this implies that G = D. This shows that the cevians AD, BE, and CF are parallel.

Case (ii). Let l be a line through A and P. We will first show that if l is not parallel to BC, then l is actually the cevian AD. To see why, let us suppose that l meets BC at G. Then the cevians AG, BE, and CF are concurrent at P, so by the first part of the theorem we must have

Since we also have

when l is known not to be parallel to BC, the proof may be completed as in case (i).

To completely finish case (ii), we must show that AP cannot be parallel to BC, and we will prove this by contradiction.

Let us suppose that AP is parallel to BC. It then follows that

and, consequently, that

Since

it follows that

But there is no point D on the line BC for which

This shows that AP cannot be parallel to BC and completes the proof of case (ii).

Proof of Menelaus’ Theorem

As with Ceva’s Theorem, we will treat the “necessary” and “sufficient” parts separately. There are a variety of different proofs for the first part, and we will give additional proofs later in this chapter.

Theorem 4.5.3. (Menelaus’ Theorem: Necessity)

If the menelaus points D, E, and F on the (extended) sides BC, CA, and AB of a triangle are collinear, then

Proof. The proof uses the fact that a transversal meets either two sides internally or no sides internally. Thus, either exactly one or all three of the ratios

are numerically negative, and consequently the sign of

is guaranteed to be negative. So, to prove the “necessary” part, we need only show that

Drop perpendiculars AR, BS, and CT from the vertices to the transversal. Then,

The reasons are

and

This completes the proof of the necessity.

For the sufficiency part of Menelaus’ Theorem, we need to prove the following:

Theorem 4.5.4. (Menelaus’ Theorem: Sufficiency)

Let D, E, and F be three menelaus points on sides BC, CA, and AB of a triangle. If

then D, E, and F are collinear.

Proof. Suppose that

Let l be the line through F and E. Then there are two possibilities:

Case (i). l meets BC at some point P.

In this case, it suffices to show that P = D. By the “necessary” part of Menelaus’ Theorem,

This, together with the hypothesis, means that

and so

implying that P = D and completing case (i).

Case (ii). l is parallel to BC.

We will show that this case cannot arise, for if l is parallel to BC, then

Now, by the hypothesis,

and since

it follows that

But it is impossible for any point D on the line BC to satisfy

This finishes the proof of the “sufficient” part of the theorem and completes the proof of Menelaus’ Theorem.

Additional Proofs

The similarity between Menelaus’ Theorem and Ceva’s Theorem is not just in the language. Here is how Menelaus’ Theorem can be used to prove the first part of Ceva’s Theorem.

Example 4.5.5. Menelaus’ Theorem implies Ceva’s Theorem.

Solution. In the figure on the right, assuming that the cevians are concurrent, our objective is to show that

The idea is to use the cevian AD of triangle ABC to decompose the triangle ABC into two “subtriangles” ABD and ADC. We then use the facts that CF is a transversal for ΔABD and BE is a transversal for ΔADC.

Applying Menelaus’ Theorem to ΔABD and transversal COF,

and also applying Menelaus’ Theorem to ΔADC and transversal BOE,

Multiplying the two equations together and cancelling a few terms, we get

Thus,

which completes the proof.

The following two proofs of Menelaus’ Theorem are from a 1937 letter that the physicist Albert Einstein wrote to the psychologist Max Wertheimer.

Einstein’s Ugly Proof

Einstein called this his “ugly proof.” It is based on similar triangles but uses a single parallel line rather than three perpendicular lines to create them.

Given ΔABC and a transversal, draw l through A parallel to the transversal hitting BC at X. Then

and the theorem follows.

Einstein’s Area Proof

Einstein called this next proof his “area proof.” It uses the following lemma.

Lemma 4.5.6. The ratio of the areas of two triangles that have a common or supplementary angle is equal to the ratio of the corresponding products of their adjacent sides.

Proof. As in the figure on the following page,

Since sin(180 − α) = sin α, the same result holds for supplementary angles.

Einstein’s Area Proof.

Given ΔABC and a transversal, then by the lemma,

from which it follows that

and the theorem follows.

4.6 Extended Versions of the Theorems

Example 4.6.1. (A Cevian Parallel to a Side)

In the figure above there are three concurrent cevians, l, m, and n, so the cevian product

should be + 1. But we have a problem: where is F? It should be the point where m meets the extended side AB, but m and AB are parallel and so that point does not exist.

We have dealt with the problem by defining the word “cevian” so that it excludes lines like m that are parallel to a side. This approach is unsatisfactory. It seems like we are playing with semantics rather than doing geometry. After all, the “cevians” in the example are concurrent, and it is a legitimate source of concern that Ceva’s Theorem cannot handle this case.

In this section, we will show that the problem can be remedied by introducing certain ideal elements called “points at infinity” or “ideal points” at which two parallel lines meet. (If you have ever looked down a long straight prairie railway track, you may have used exactly the same language and said that the rails “meet at infinity.”)

When we introduce ideal points, we must be careful that we do not create more problems than we remove. The following conventions and definitions will help us avoid complications.

We append to the plane a collection of ideal points or points at infinity. This collection of ideal points is called the ideal line or the line at infinity. Points that are not ideal points are called ordinary points, lines that are not ideal lines are called ordinary lines. (The words “point” and “line” by themselves do not distinguish between the ordinary and the ideal versions.)

We postulate that:

The plane together with all of the ideal points is called the extended Euclidean plane or simply the extended plane. The plane without the ideal line is called the Euclidean plane to distinguish it from the extended plane.³

We designate an ideal point by an arrow indicating the direction in which it lies, and an arrow pointing in exactly the opposite direction designates the same ideal point. In the figure below, parallel lines l₁, l₂, and l₃ all meet at the ideal point I. The lines m₁ and m₂ meet at a different ideal point J.

In the extended plane, every two distinct points determine a unique line, and every two distinct lines determine a unique point. Technically, in the extended plane there are no parallel lines. Nevertheless, we will continue to use the word parallel to describe ordinary lines in the extended plane that meet only at an ideal point.

If I is the ideal point on the line AB, we cannot assign a real number to the directed distance AI. However, Ceva’s Theorem deals with directed ratios, and we adopt the convention that

for the ideal point on the line AB.

This makes a certain amount of sense, for two reasons:

4.6.1 Ceva’s Theorem in the Extended Plane

In the extended Euclidean plane, a cevian is defined to be any line that contains exactly one vertex of an ordinary triangle. (An ordinary triangle is one whose edge segments do not contain ideal points.) Thus, the notion of a cevian has been modified to include lines through a vertex that in the Euclidean plane would be parallel to the opposite side.

The following two examples show that Ceva’s Theorem is valid in the extended Euclidean plane with the modified notion of a cevian.

Example 4.6.2. Given an ordinary triangle ABC and three concurrent cevians l, m, and n, with m parallel to BC, show that the cevian product is 1.

Solution. We have

and by similar triangles,

showing that the cevian product is 1.

Example 4.6.3. Given ΔABC and three concurrent cevians l, m, and n, with l parallel to BC and n parallel to AB, show that the cevian product is 1.

Solution.

since the diagonals of parallelogram ABCP bisect each other.

The previous two examples show that in the extended plane. Ceva’s Theorem can be stated more succinctly.

Theorem 4.6.4. (Ceva’s Theorem for the Extended Plane)

A necessary and sufficient condition that the cevians AD, BE, and CF of a triangle ABC be concurrent is that

The theorem includes the possibility that the point of concurrency is an ideal point; that is, it includes the possibility that the three cevians are parallel in the Euclidean plane. It also includes the possibility that some of the vertices are ideal points, but one must be careful about the sufficiency part of the theorem.

4.6.2 Menelaus’ Theorem in the Extended Plane

In the extended Euclidean plane, the notion of a transversal is modified to include those that are parallel to a side of an ordinary triangle: any ordinary line that does not contain a vertex of the triangle is called a transversal of the triangle.

Example 4.6.5. Show that if a transversal l is parallel to side BC of an ordinary-triangle ABC, then the corresponding cevian product is − 1.

Solution. Let us suppose that l meets sides BC, CA, and AB at D, E, and F, respectively. Then D is an ideal point and, since EF is parallel to BC,

so that

An alternate proof: drop perpendiculars from the vertices, as in the figure on the right. Then

since .

In the extended Euclidean plane, the statement of Menelaus’ Theorem is the same as before.

Theorem 4.6.6. (Menelaus’ Theorem for the Extended Plane)

Let ABC be an ordinary triangle, and let D, E, F be nonvertex points on the (possibly extended) sides BC, CA, and AB of the triangle. Then the points D, E, and F are collinear if and only if

Although the statement is the same, the meaning is a bit different, for in the extended version neither the vertices nor the menelaus points are restricted to being ordinary points.

Remark. The increased generality of the theorems does have some cost. If we wish to apply the extended versions to Euclidean problems, we must first restate the problem in the extended plane. For example, the theorem about the concurrency of the medians of a triangle would, in the extended plane, be stated as follows:

To prove this by the extended version of Ceva’s Theorem, we would have to show that the point of concurrency is not an ideal point; that is, we still have to show that the cevians in question are not parallel.

4.7 Problems

² The word “cevian” is a combination of Ceva’s name and the word “median.”

³ The extended plane is also called the projective plane and is the setting for what is called “projective geometry” in Part III.