Ceva's theorem
- Olympiad · IOQM
Ceva's theorem gives a test for when three lines, each from a corner of a triangle to the opposite side, all pass through one point.
Take a triangle ABC. Mark D on side BC, E on side CA and F on side AB. Draw the lines AD, BE and CF. Ceva's theorem says these three lines meet at one point exactly when:
- (BD/DC) × (CE/EA) × (AF/FB) = 1.
Lines that meet at one point are called concurrent. To remember the order, walk round the triangle: B to D to C, then C to E to A, then A to F to B. The theorem is named after Giovanni Ceva, an Italian mathematician.
A small example
A median joins a corner to the mid-point of the opposite side. So for the three medians, BD = DC, CE = EA and AF = FB. Each ratio is 1, and 1 × 1 × 1 = 1. By Ceva's theorem, the three medians meet at one point. That point is the centroid.
Now suppose BD/DC = 2 and CE/EA = 1/2. Where must F be for the lines to meet? We need 2 × 1/2 × (AF/FB) = 1. So AF/FB = 1, and F is the mid-point of AB. The ratios here are the same kind you meet with the section formula.
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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