Ceva's and Menelaus's Theorems

The line segment joining a vertex of a triangle to any given point on the opposite side is called a cevian. Thus, if X, Y and Z are points on the respective sides BC, CA and AB of triangle ABC, the segments AX, BY and CZ are cevians. This term comes from the name of the Italian mathematician Giovanni Ceva, who published in 1678 the following very useful theorem:

Ceva's Theorem If three cevians AX, BY and CZ, one through each vertex of a triangle ABC, are concurrent, then

$$\frac{BX}{XC}\frac{CY}{YA}\frac{AZ}{ZB}=1.$$

Conversely, if this equation holds for points X, Y and Z on the three sides, then these three point are concurrent. (We say that three lines or segments are concurrent if they all pass through one point)

  

Figure 2: Ceva's theorem

Proof. Given the concurrence we can use that the areas of the triangles with equal altitudes are proportional to the bases of the triangles. Referring to Figure 2, we have

$$\frac{BX}{XC}=\frac{S_{ABX}}{S_{AXC}}=\frac{S_{PBX}}{S_{PXC}}=\frac{S_{ABX}-S_{PBX}}{S_{AXC}-S_{PXC}}=\frac{S_{ABP}}{S_{CAP}}.$$

Similarly,

$$\frac{CY}{YA}=\frac{S_{BCP}}{S_{ABP}}; \frac{AZ}{ZB}=\frac{S_{CAP}}{S_{BCP}}.$$

Now, if we multiply these, we find

$$\frac{BX}{XC}\frac{CY}{YA}\frac{AZ}{ZB}=\frac{S_{ABP}}{S_{CAP}}\frac{S_{BCP}}{S_{ABP}}\frac{S_{CAP}}{S_{BCP}}=1.$$

Conversely, suppose that the first two cevians meet at P, as before, and that the third cevian through this point P is CZ'. Then, by the just proved part of the theorem, we have

$$\frac{BX}{XC}\frac{CY}{YA}\frac{AZ'}{Z'B}=1$$

.

But we are assuming

$$\frac{BX}{XC}\frac{CY}{YA}\frac{AZ}{ZB}=1.$$

Hence

$$\frac{AZ'}{Z'B}=\frac{AZ}{ZB}$$

and adding 1 at both sides we obtain that Z' coincides with Z, and thus AX, BY and CZ are concurrent.

Now is not an problem to prove many famous theorems stating that certain cevians have common point, for example that medians (altitude, internal bisectors) are concurrent. One more property of this type is the following:

Problem 2. The cevians AX, BY and CZ are concurrent, where X, Y and Z are as before the points where the incircle touches the sides BC, CA and AB of $\Delta ABC$ (The common point of these cevians is called Gergonne point of $\Delta ABC$).

Proof. This is immediate corollary of Theorem 1 of the lecture ``Advanced Geometry - 2'' and Ceva's theorem.

Menelaus of Alexandria (about 100 A.D. , not to be confused with Menelaus of Sparta) wrote a treatise called Sphaerica in which he used a certain property of a spherical triangle. May be it was; but since no earlier record of it has survived, we shall simply call the assertion of this property Menelaus's theorem. In the notation of Figure 3 it may be stated as follows:

Menelaus's Theorem. If three points X, Y and Z on the sides BC, CA, AB (suitably extended) of $\Delta ABC$ are collinear, the

$$\frac{BX}{XC}\frac{CY}{YA}\frac{AZ}{ZB}=1.$$

Conversely, if this equation holds for points X, Y and Z on the three sides, such that either any one of them or all three lie outside the corresponding segments, then these three point are collinear. (We say that three points are collinear if they all are on the same line)

  

Figure 3: Menelaus's theorem

Proof. Given the collinearity of X, Y and Z, as in Figure 3, let h₁, h₂ and h₃ be the lengths of the perpendiculars from A, B and C to the line XY. From the three equations

$$\frac{BX}{CX}=\frac{h_2}{h_3},\frac{CY}{AY}=\frac{h_3}{h_1},\frac{AZ}{BZ}=\frac{h_1}{h_2},$$

we obtain the desired result by multiplication. (Notice that that the same holds in the case when just one of the sides of $\Delta ABC$ must be extended to accommodate the three distinct points, X, Y and Z, see Figure 4.)

  

Figure 4: Menelaus's Theorem

Conversely, if X, Y and Z occur in the described in the theorem way and

$$\frac{BX}{XC}\frac{CY}{YA}\frac{AZ}{ZB}=1,$$

let the line AB and XY meet at Z'. Then

$$\frac{BX}{XC}\frac{CY}{YA}\frac{AZ'}{Z'B}=1.$$

Hence

$$\frac{AZ}{ZB}=\frac{AZ'}{Z'B},$$

so Z' coincides with Z and thus X, Y and Z are collinear.

Menelaus's theorem provides a criterion for collinearity, just as Ceva's theorem. (To emphasize the contrast, see the note below.)

Note. The Menelaus's theorem may be stated in the terms of directed segments. For the teachers it is good to express the theorem in the alternative form using the directed segments:

$$\frac{\overline{BX}}{\overline{XC}}\frac{\overline{CY}}{\overline{YA}}\frac{\overline{AZ}}{\overline{ZB}}=-1.$$

Remind you that for three points M, N and L lying on the same line, the fraction $\frac{\overline{MN}}{\overline{NL}}$ has positive value if N is between M and L and negative otherwise.

We can obtain a strait-forward application of Menelaus's theorem, using the properties of the internal and external bisectors:

Problem 3. The internal bisectors of two angles of a given triangle, and the external bisector of the third angle, meet their respective sides at three collinear points (see Figure 5).

  

Figure 5: problem 3

Proof. Lets denote AU, BV and CW the corresponding bisectors. From the well-known property of the internal and external bisectors we have that:

$$\frac{CU}{BU}=\frac{AC}{AB},\frac{CV}{AV}=\frac{BC}{AC}, \frac{AW}{BW}=\frac{AC}{BC},$$

so we can apply the Menelaus's theorem.