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Menelaus's Theorem — Definition, Formula & Examples

Menelaus’s Theorem
Theorem of Menelaus

A theorem relating the way two cevians of a triangle divide each other and two of the triangle's sides.

Triangle ABC crossed by a terracotta transversal that meets sides AB at D, AC at F, and the extension of BC at P; below: (AD/DB)·(BP/PF)·(FC/CA) = 1.

Key Formula

AD‾DB‾⋅BE‾EC‾⋅CF‾FA‾=−1\frac{\overline{AD}}{\overline{DB}} \cdot \frac{\overline{BE}}{\overline{EC}} \cdot \frac{\overline{CF}}{\overline{FA}} = -1
Where:
  • △ABC\triangle ABC = The given triangle with vertices A, B, and C.
  • DD = The point where the transversal meets side AB (or its extension).
  • EE = The point where the transversal meets side BC (or its extension).
  • FF = The point where the transversal meets side CA (or its extension).
  • AD‾DB‾\frac{\overline{AD}}{\overline{DB}} = The signed (directed) ratio in which D divides segment AB.
  • −1-1 = The product equals −1 (using signed ratios). When using unsigned lengths, the product equals 1, and an odd number of the points D, E, F must lie on the extensions of the sides rather than on the sides themselves.

Worked Example

Problem: In triangle ABC, a straight line crosses side AB at D, side BC at E, and the extension of side CA at F. You are given AD = 2, DB = 3, BE = 4, and EC = 6. Verify that D, E, and F are collinear using Menelaus's Theorem and find CF/FA.
Step 1: Write Menelaus's Theorem using unsigned lengths. Since exactly one point (F) lies on an extension rather than on the side itself, the unsigned form of the theorem states the product of the three ratios equals 1.
ADDB⋅BEEC⋅CFFA=1\frac{AD}{DB} \cdot \frac{BE}{EC} \cdot \frac{CF}{FA} = 1
Step 2: Substitute the known values for the first two ratios.
23⋅46⋅CFFA=1\frac{2}{3} \cdot \frac{4}{6} \cdot \frac{CF}{FA} = 1
Step 3: Simplify the product of the first two fractions.
23⋅46=818=49\frac{2}{3} \cdot \frac{4}{6} = \frac{8}{18} = \frac{4}{9}
Step 4: Solve for CF/FA by dividing both sides by 4/9.
CFFA=149=94\frac{CF}{FA} = \frac{1}{\frac{4}{9}} = \frac{9}{4}
Step 5: Interpret the result: F divides the extension of side CA such that CF/FA = 9/4. Since the product of the three unsigned ratios equals 1 and exactly one of the three points lies outside its side, Menelaus's condition is satisfied, confirming D, E, F are collinear.
Answer: CF/FA = 9/4, and the three points are collinear, consistent with Menelaus's Theorem.

Another Example

This example starts from the collinearity condition and solves for an unknown ratio. It also shows why Menelaus's Theorem often distinguishes between signed ratios and unsigned lengths.

Problem: In triangle PQR, a transversal meets side QR at X, side RP at Y, and the extension of side PQ at Z. Given QX = 5, XR = 3, RY = 6, and YP = 4, find the magnitude of PZ/ZQ.
Step 1: Because Z is on an extension, use the unsigned form of Menelaus's Theorem. With one external point, the product of the three positive ratios is 1.
QXXR⋅RYYP⋅PZZQ=1\frac{QX}{XR} \cdot \frac{RY}{YP} \cdot \frac{PZ}{ZQ} = 1
Step 2: Substitute the known lengths.
53⋅64⋅PZZQ=1\frac{5}{3} \cdot \frac{6}{4} \cdot \frac{PZ}{ZQ} = 1
Step 3: Simplify the product of the first two ratios.
53⋅64=3012=52\frac{5}{3} \cdot \frac{6}{4} = \frac{30}{12} = \frac{5}{2}
Step 4: Solve for the remaining ratio.
52⋅PZZQ=1  ⟹  PZZQ=25\frac{5}{2} \cdot \frac{PZ}{ZQ}=1 \implies \frac{PZ}{ZQ}=\frac{2}{5}
Step 5: Interpret the result. The required magnitude is 2/5; in the signed form, the corresponding ratio would carry a negative sign because Z lies on an extension.
∣PZZQ∣=25\left|\frac{PZ}{ZQ}\right|=\frac{2}{5}
Answer: The magnitude of PZ/ZQ is 2/5.

Frequently Asked Questions

What is the difference between Menelaus's Theorem and Ceva's Theorem?
Both theorems involve ratios along the sides of a triangle, but they apply to different configurations. Ceva's Theorem deals with three cevians (lines from each vertex to the opposite side) that are concurrent (meet at a single point), and the product of the unsigned ratios equals 1. Menelaus's Theorem deals with a single transversal line that crosses the three sides (or their extensions), and the product of the signed ratios equals −1. The sign difference reflects the geometric distinction between concurrence and collinearity.
Why does Menelaus's Theorem use signed ratios?
Signed (directed) ratios account for whether a dividing point lies between the two vertices of a side or on the extension beyond one vertex. If D lies between A and B, the ratio AD/DB is positive; if D lies outside segment AB, the ratio is negative. The signed ratio convention ensures the theorem gives a clean result of exactly −1, and it automatically encodes the geometric constraint that an odd number of the three points must lie on extensions of the triangle's sides.
When do you use Menelaus's Theorem?
Use Menelaus's Theorem when you need to prove that three points on the sides (or extensions) of a triangle are collinear, or when you know three points are collinear and want to find an unknown ratio. It appears frequently in olympiad geometry, projective geometry, and problems involving transversals cutting across triangles.

Menelaus's Theorem vs. Ceva's Theorem

Menelaus's TheoremCeva's Theorem
ConfigurationA transversal line crossing the three sides of a triangleThree cevians drawn from each vertex to the opposite side
Key conditionThe three intersection points are collinearThe three cevians are concurrent (meet at one point)
Formula (signed ratios)(AD/DB)·(BE/EC)·(CF/FA) = −1(AD/DB)·(BE/EC)·(CF/FA) = +1
Points on extensionsAn odd number (1 or 3) of points lie on side extensionsAn even number (0 or 2) of points lie on side extensions
Typical useProving collinearity or finding ratios along a transversalProving concurrency or finding ratios along cevians

Why It Matters

Menelaus's Theorem is one of the foundational tools in triangle geometry, appearing regularly in mathematical olympiads and competitions. It provides the standard technique for proving that three points are collinear, complementing Ceva's Theorem for concurrence. You will also encounter it in projective geometry and in more advanced courses that study cross-ratios and harmonic divisions.

Common Mistakes

Mistake: Using unsigned ratios and expecting the product to equal −1.
Correction: When you use unsigned (absolute-value) ratios, the product equals 1, not −1. The −1 result applies only when you use signed (directed) ratios, where the sign indicates whether the point lies between the two vertices or on an extension. Be consistent: either use signed ratios throughout or use unsigned ratios and separately verify the odd-external-point condition.
Mistake: Traversing the sides in an inconsistent order.
Correction: You must walk around the triangle in a consistent cyclic direction (e.g., A→B→C→A) when forming each ratio. If D is on AB, the ratio should be AD/DB (from A toward B). Swapping numerator and denominator for one ratio changes the product and leads to an incorrect conclusion.

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