Menelaus' theorem
- Olympiad · IOQM
Menelaus' theorem gives a test for when three points, one on each side line of a triangle, lie on one straight line.
Take a triangle ABC. A straight line cuts the line BC at D, the line CA at E and the line AB at F. At least one of these points lies outside its side, on the side made longer. Menelaus' theorem says:
- (BD/DC) × (CE/EA) × (AF/FB) = 1, using lengths.
It is a partner to Ceva's theorem, with the same product. Ceva tests whether three lines meet at one point. Menelaus tests whether three points lie on one line. Such points are called collinear. With signed lengths the product is -1, and then the test works both ways.
A small example
In triangle ABC, F is the mid-point of AB, so AF/FB = 1. E is on CA with CE/EA = 2. The line through F and E meets line BC at D. Where is D?
- By the theorem, (BD/DC) × 2 × 1 = 1, so BD/DC = 1/2.
- So DC is twice BD. A line crosses at most two sides inside the triangle, and F and E use both. So D is outside BC.
- D lies beyond B, with DC = DB + BC. Then 2BD = BD + BC, so BD = BC.
So D is as far past B as C is from B. You can check this with coordinates.
Where it fits
This goes beyond the Class 10 board syllabus. HBCSE's syllabus for the Mathematical Olympiad names it under plane geometry. The Mathematics Teachers' Association (India), MTA(I), conducts IOQM, the first stage of the Mathematical Olympiad Programme that HBCSE organises for NBHM.
Sources
Facts last checked against these sources on 30 September 2026.
- Syllabus for Mathematical Olympiad (Plane Geometry) · HBCSE, TIFR
- Mathematical Olympiad 2026-2027: stages of selection · HBCSE, TIFR
- Brochure: Mathematical Olympiads 2026-2027 · HBCSE, TIFR
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