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AMC 12 focus

Mass Points, Ceva & Menelaus

Turning ratio-and-concurrency problems into arithmetic — assign weights, or apply one product identity.

Mass points

Mass points assign a “weight” to each vertex of a triangle so that a cevian's foot balances like a lever: if a segment from BB to a point DD on ACAC divides it so that AD:DC=m:nAD:DC = m:n, assign mass nn to AA and mass mm toCC — then DD's mass is the sum, and it “balances” along ACAC. Chaining this across multiple cevians turns a ratio-finding problem into simple weighted averaging, with no coordinates required.

Ceva's theorem

Three cevians AD,BE,CFAD, BE, CF (one from each vertex, to a point on the opposite side) are concurrent (all pass through one point) if and only if:

AFFBBDDCCEEA=1\dfrac{AF}{FB}\cdot\dfrac{BD}{DC}\cdot\dfrac{CE}{EA} = 1

Menelaus's theorem

Menelaus's theorem uses the exact same product, but for a very different setup: a single transversal line crossing the (possibly extended) three sides of a triangle at points D,E,FD, E, F. Those three points are collinear if and only if:

AFFBBDDCCEEA=1\dfrac{AF}{FB}\cdot\dfrac{BD}{DC}\cdot\dfrac{CE}{EA} = 1

The two theorems are easy to mix up because the formula is identical — the difference is entirely in what's being tested: three concurrent cevians (Ceva) versus three collinear pointson the extended sides (Menelaus).

Worked example

Problem: In ABC\triangle ABC, cevians AD,BE,CFAD, BE, CF are concurrent, with AFFB=23\tfrac{AF}{FB}=\tfrac23 and BDDC=34\tfrac{BD}{DC}=\tfrac34. FindCEEA\tfrac{CE}{EA}.

Solution: By Ceva's theorem:

2334CEEA=1    12CEEA=1    CEEA=2\dfrac23 \cdot \dfrac34 \cdot \dfrac{CE}{EA} = 1 \;\Longrightarrow\; \dfrac12 \cdot \dfrac{CE}{EA} = 1 \;\Longrightarrow\; \dfrac{CE}{EA}=2

Practice problems

1.Cevians are concurrent with AFFB=1\tfrac{AF}{FB}=1, BDDC=2\tfrac{BD}{DC}=2. Find CEEA\tfrac{CE}{EA}.

2.A median, by definition, has its foot at the midpoint of a side. Explain why the three medians automatically satisfy Ceva's theorem.

3.A transversal crosses the sides (or extensions) of a triangle with AFFB=32\tfrac{AF}{FB}=\tfrac32 and BDDC=13\tfrac{BD}{DC}=\tfrac13. Find CEEA\tfrac{CE}{EA} for the points to be collinear (Menelaus).

4.What is the key difference between when you'd apply Ceva's theorem versus Menelaus's theorem?