The theorems by Menelaus and Ceva

Menelaus's theorem says the following:

Consider a triangle with vertices \(A\), \(B\) and \(C\). Let \(D\) be a point on the line \(\langle A, B\rangle\), \(E\) a point \(\langle B, C\rangle\) and \(F\) a point on \(\langle A, C\rangle\). Then the points \(D\), \(E\) and \(F\) are collinear if and only if

\begin{align} \frac{AF}{FC} \cdot \frac{CE}{EB} \cdot \frac{BD}{DA} = -1. \label{eq:menelaus} \end{align}

Here the lengths of the segments are taken with orientation, so \(AB\) is positive if the point \(B\) is to the right of point \(A\) (after fixing some orientation of the line through \(A\) and \(B\)). The following graphic shows the Menelaus configuration:

One way to remember the orientation of the segments in equation \eqref{eq:menelaus} is like this: Start at a vertex of the triangle, say \(A\), and jump to one of the intersection points of the red line with one of the sides through \(A\), for example \(F\). This gives the segment \(AF\). From there, jump to the other vertex on the same line, \(C\), and get the segment \(FC\). Then continue in this way, so jump from \(C\) to the intersection point \(E\), then from \(E\) to the vertex \(B\) and so on.

The theorem is easy to proof using similar triangles as we will see. It is needed for the proof of Desargues' theorem purely in the plane, i.e. without lifting the Desargues configuration to three dimensions.

Here is how to prove the theorem:

Let the points \(D\), \(E\) and \(F\) be collinear on the line \(\ell\). Let \(A'\), \(B'\) and \(C'\) be the orthogonal projections of the vertices \(A\), \(B\) and \(C\) onto the line \(\ell\). The triangles \(D A A'\) and \(D B B'\) have the same angles so they are similar triangles. Consequently the side lengths satisfy

\begin{align} \left|\frac{BD}{DA}\right| = \left|\frac{BB'}{AA'}\right|. \end{align}

Likewise, the triangles \(E B B'\) and \(E C C'\) are similar, so

\begin{align} \left|\frac{CE}{EB}\right| = \left|\frac{CC'}{BB'}\right|. \end{align}

The last pair of similar triangles, \(F A A'\) and \(F C C'\), gives us

\begin{align} \left|\frac{AF}{FC}\right| = \left|\frac{AA'}{CC'}\right|. \end{align}

Multiplying the three equations, we get the statement of the theorem up to the sign. The left-hand side of the equation is always negative as either exactly one or all three of the ratios are negative. The former case happens if the line through \(D\), \(E\) and \(F\) meets the triangle \(ABC\) in two points; the latter case happens if it misses the triangle.

For the other direction, let the points \(D\), \(E\) and \(F\) be given such equation \eqref{eq:menelaus} holds. We want to show that \(D\), \(E\) and \(F\) are collinear. Consider the line through \(E\) and \(F\). This line intersects \(\langle A, B\rangle\) in a point \(D'\). As \(D'\), \(E\) and \(F\) are collinear, the other direction of Menelaus's theorem can be applied and the equation from his theorem also holds with \(D\) replaced by \(D'\). Comparing the two equations it follows that

\begin{align} \left|\frac{BD'}{D'A}\right| = \left|\frac{BD}{DA}\right|, \end{align}

so \(D = D'\).¹

Menelaus's theorem is about three collinear points. Ceva's theorem on the other hand is about three concurrent lines.

Consider a triangle with vertices \(A\), \(B\) and \(C\). Let \(D\) be a point on the line \(\langle A, B\rangle\), \(E\) a point on \(\langle B, C\rangle\) and \(F\) a point on \(\langle A, C\rangle\). Then the lines \(\langle A, E\rangle\), \(\langle C, D\rangle\) and \(\langle B, F\rangle\) are concurrent in a point \(P\) if and only if

\begin{align} \frac{AF}{FC} \cdot \frac{CE}{EB} \cdot \frac{BD}{DA} = 1. \label{eq:ceva} \end{align}

Note that the sign on the right hand side is different from Menelaus's theorem. Here is a picture of the configuration:

The standard proof uses the area of triangles:

The triangles \(APF\) and \(CPF\) share the same height and so do the triangles \(ABF\) and \(CBF\). Writing \(|ABC|\) for the area of a triangle with vertices \(A\), \(B\) and \(C\) we therefore have

\begin{align} \frac{|APF|}{|CPF|} = \frac{|ABF|}{|CBF|} = \frac{AF}{FC} \end{align}

Rearranging the equation gives

\begin{align} \frac{AF}{FC} = \frac{|ABF| - |APF|}{|CBF| - |CPF|} = \frac{|BAP|}{|CBP|} \end{align}

and similarly

\begin{align} \frac{CE}{EB} = \frac{|ACP|}{|BAP|} \end{align}

and

\begin{align} \frac{BD}{DA} = \frac{|CBP|}{|ACP|}. \end{align}

Multiplying the last three equations we get the "only if" direction of the theorem.

The other direction can be done in analogy to Menelaus's theorem. Let \(D\), \(E\) and \(F\) in the theorem be given such that the equation holds and let \(P\) be the point where the lines \(\langle A, E\rangle\) and \(\langle C, D\rangle\) meet. The line \(\langle B, P\rangle\) then intersects the line \(\langle A, C\rangle\) in a point \(F'\) and the only if direction of Ceva's theorem holds for the triangle \(ABC\) and the points \(D\), \(E\) and \(F'\). It follows that

\begin{align} \frac{AF}{FC} = \frac{AF'}{F'C} \end{align}

so \(F = F'\).

In this paper a more unified proof which proves both theorems from the same configuration of points and lines. The configuration is the following:

The proof uses projective geometry, so a key player is the cross-ratio of four collinear points \(A\), \(B\), \(C\) and \(D\) defined as

\begin{align} [A, B, C, D] = \frac{AC \cdot BD}{AD \cdot BC}. \label{eq:cross-ratio-def} \end{align}

This quantity is invariant under projections and projecting the points \(X\), \(B\), \(C\) and \(D\) from \(Q\) onto the lines \(\langle A, B\rangle\) and \(\langle A, C\rangle\) respectively we obtain following equalities:

\begin{align} [X, B, C, D] = [A, B, Z', F], \quad [X, C, D, B] = [A, C, E, Y']. \end{align}

Similarly, projecting the points \(P\), \(Q\), \(A\) and \(X\) from \(C\) onto the line \(\langle A, B\rangle\) and from \(B\) onto the line \(\langle A, C\rangle\) we also get the following equality:

\begin{align} [Y', Y, A, C] = [Z', Z, A, B]. \end{align}

Moreover, one can check from the definition \eqref{eq:cross-ratio-def} that the cross satisfies the equation

\begin{align} [X, D, B, C] \cdot [X, B, C, D] \cdot [X, C, D, B] = -1. \end{align}

Using the equalities from the projectivities above, we get:

\begin{align} \begin{split} -1 &= [X, D, B, C] \cdot [A, B, Z', F] \cdot [A, C, E, Y'] \\ &= \frac{XB \cdot DC}{XC \cdot DB} \cdot \frac{AZ' \cdot BF}{AF \cdot BZ'} \cdot \frac{AE \cdot CY'}{AY' \cdot CE} \\ &= \frac{XB \cdot YC \cdot ZA}{XC \cdot YA \cdot ZB} \cdot \frac{DC \cdot BF \cdot AE}{DB \cdot AF \cdot CE} \end{split} \label{eq:ceva-proof-1} \end{align}

Sending the points \(D\), \(E\) and \(F\) to the line at infinity, the second factor becomes one and we get

\begin{align} \frac{BX}{XC} \cdot \frac{CY}{YA} \cdot \frac{AZ}{ZB} = 1 \end{align}

which is Ceva's theorem.

Menelaus's theorem can be deduced from this. Consider the triangle \(ABC\). Because the lines \(\langle B, Y'\rangle\), \(\langle C, Z'\rangle\) and \(\langle A, X\rangle\) meet in \(Q\), Ceva's theorem tell us that

\begin{align} \frac{BX}{XC} \cdot \frac{CY'}{Y'A} \cdot \frac{AZ'}{Z'B} = 1. \end{align}

Using this relation in the second line of equation \eqref{eq:ceva-proof-1} we immediately get

\begin{align} \frac{AE}{CE} \cdot \frac{CD}{DB} \cdot \frac{BF}{FA} = 1. \end{align}

• H.G. Green: On the Theorems of Ceva and Menelaus