Ceva or Menelaus: when to use which

Ceva and Menelaus are the two theorems students most often mix up, and the reason is that they look identical on the page. Both take three points, one on each side of a triangle. Both multiply three ratios. Both say the answer is $1$. If you have only ever seen them written down, there is genuinely nothing in the formula to tell you which is which.

The formula is not where the difference lives. The configuration is. This page is about reading the configuration.

The one question

Before anything else, ask what the problem wants:

• Do three lines pass through a single point? That is Ceva.
• Do three points lie on a single line? That is Menelaus.

Concurrency is Ceva. Collinearity is Menelaus. If the problem states which of those it is after, you are already done choosing, and everything below is for the cases where it does not.

The reason this works is that the two theorems are answers to dual questions. Ceva's theorem says the cevians $AD$, $BE$, $CF$ meet at one point exactly when

$$\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = 1$$

and Menelaus' theorem says that same product is $1$ when $D$, $E$, $F$ instead lie on one straight line. Two of geometry's oldest questions, reduced to the same multiplication.

When the problem does not say

Half the time you are not asked to prove concurrency or collinearity at all. You are given some ratios and asked for another one, and the theorem is a tool for a computation rather than the conclusion. Then you have to read the picture, and there is a counting rule that does it reliably.

That rule is exact, and it is worth understanding rather than memorising. Three cevians through an interior point cut all three sides internally: zero outside, even. Move the concurrency point outside the triangle and exactly two of the three feet leave their segments: still even. A straight line, on the other hand, can never cut all three sides of a triangle internally. It enters and it leaves, which uses up two sides, so it must meet the third side's extension: one outside, odd.

So the parity is not a coincidence to be learned. It is the difference between a point, which can sit anywhere, and a line, which has to get out of the triangle somehow.

A worked Ceva

In triangle $ABC$, take $D$ on $BC$ with $BD/DC = 2$ and $E$ on $CA$ with $CE/EA = 3$. Where must $F$ sit on $AB$ for $AD$, $BE$, $CF$ to be concurrent?

Ceva answers it in one line: $2 \cdot 3 \cdot \frac{AF}{FB} = 1$, so $AF/FB = \tfrac16$. Nothing about the shape of the triangle enters, which is the real content of the theorem. The concurrency is a fact about ratios only.

Now notice what Ceva does not tell you. It says the cevians meet. It says nothing about where on $AD$ they meet, and problems ask that constantly. For that, switch interfaces.

Put masses at the vertices so each foot balances: $BD/DC = 2$ needs $m_B : m_C = 1 : 2$, and $CE/EA = 3$ needs $m_C : m_A = 1 : 3$. Take $m_C = 2$ and the three masses are $m_A = 6$, $m_B = 1$, $m_C = 2$. Then $D$ carries $m_B + m_C = 3$, and the concurrency point $P$ splits $AD$ in the inverse ratio of the masses at its ends: $AP/PD = 3/6 = \tfrac12$.

Mass points are Ceva with the answer to "where" included. On a timed paper that is usually the version you want.

A worked Menelaus

In triangle $ABC$, let $M$ be the midpoint of $BC$ and let $P$ on $AB$ satisfy $AP/PB = \tfrac13$. Line $PM$ meets line $AC$ at $Q$. Find $AQ/QC$.

Apply Menelaus to triangle $ABC$ with the transversal through $P$, $M$, $Q$:

$$\frac{AP}{PB}\cdot\frac{BM}{MC}\cdot\frac{CQ}{QA} = 1$$

which gives $\tfrac13 \cdot 1 \cdot \frac{CQ}{QA} = 1$, so $CQ/QA = 3$ and $AQ/QC = \tfrac13$.

Check the parity as a sanity test. $P$ is inside $AB$, $M$ is inside $BC$, and $Q$ turns out to lie beyond $A$ on the extension of $CA$: one outside, odd, Menelaus. Had you counted before computing, the rule would have chosen the theorem for you.

The unification, and why it is worth the extra step

Both products above came out to $1$, which is exactly the ambiguity this page opened on. Signed ratios remove it. Give $BD/DC$ a sign, positive when $D$ is inside $BC$ and negative when it is outside, and the two theorems become one statement with two values:

• concurrent cevians give a signed product of $+1$;
• collinear points give a signed product of $-1$.

The parity rule is that sign, spelled out in whole numbers: an odd number of external division points is an odd number of minus signs, which is what turns $+1$ into $-1$. Learning the signed version means you carry one theorem instead of two, and you stop guessing.

Two configurations that pass the smell test and still are not these theorems

Recognition is as much about ruling out as ruling in, and both theorems have a hard requirement that is easy to skim past: three points, one on each side-line of one triangle.

The medians are concurrent is Ceva at its cleanest. Each midpoint gives a ratio of $1$, the product is $1 \cdot 1 \cdot 1$, and the concurrency follows immediately. Mass points then add what Ceva cannot see: equal masses at the vertices put mass $2$ at each midpoint, so the centroid splits every median $2 : 1$ from the vertex.

The perpendicular bisectors are concurrent is a concurrency, and Ceva does not touch it. The perpendicular bisector of $AB$ passes through the midpoint of $AB$, but it does not pass through $C$, so it is not a cevian at all. There are no vertex-to-side lines here, so there is no ratio product to write down. The proof is the distance characterisation instead: a point is equidistant from $A$ and $B$ exactly when it is on their perpendicular bisector.

The Euler line is the mirror image of that mistake. It is a collinearity, and Menelaus still does not apply, because $H$, $G$ and $O$ are not three points sitting one on each side-line of a triangle. Vectors with the circumcentre at the origin settle it in four lines.

The test to run before writing either product is small and it saves whole minutes: name the triangle, then name which side-line each of your three points is on. If you cannot fill in all three, you have the wrong tool.

What to practise

Do not practise the theorems. Practise the classification. Take a stack of triangle problems and, for each one, write three lines only: which triangle, which three points, concurrency or collinearity. Then check the parity of the external points against your answer. You will do twenty of those in the time one full solution takes, and the choosing is the part you are actually short of.

Lemma teaches Ceva, Menelaus and mass points as one named technique for exactly this reason, with the signed form and the mass-point interface attached rather than filed separately. The problem archive has the classical configurations above in full, with difficulty and expected time on each, and the daily problem posts a fresh one every day in three tiers with no account needed.