Ceva's Theorem Calculator
Check if three cevians in a triangle are concurrent
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Understanding Ceva's Theorem
Ceva's theorem (named after Italian mathematician Giovanni Ceva, 1647-1734) provides a necessary and sufficient condition for three cevians of a triangle to be concurrent (meet at a single point).
The Theorem
Cevians AD, BE, and CF are concurrent (meet at point P) if and only if the product of the three ratios equals 1.
Important Special Cases
Medians (Centroid)
When F, D, E are midpoints: AF/FB = BD/DC = CE/EA = 1, so product = 1. Medians meet at the centroid.
Altitudes (Orthocenter)
Altitudes are perpendicular cevians. They satisfy Ceva's theorem and meet at the orthocenter.
Angle Bisectors (Incenter)
By the angle bisector theorem, the ratios depend on adjacent sides. They meet at the incenter.
Relationship to Menelaus
Ceva's and Menelaus's theorems are dual to each other:
- Ceva: Checks concurrency (cevians meet at a point)
- Menelaus: Checks collinearity (points lie on a line)
- Both use the same ratio formula but test different geometric properties
Frequently Asked Questions
What does concurrent mean?
Concurrent means that three or more lines (or line segments) meet at a single point. Ceva's theorem helps verify this condition for cevians in a triangle.
Who was Giovanni Ceva?
Giovanni Ceva (1647-1734) was an Italian mathematician who proved this theorem in 1678 in his work "De lineis rectis." He also made contributions to hydraulics and economics.
What is a cevian point?
A cevian point is the point where three concurrent cevians meet. Famous cevian points include the centroid, orthocenter, incenter, and circumcenter.
Do all triangles have concurrent cevians?
Yes! Every triangle has several sets of concurrent cevians: medians (meeting at centroid), altitudes (orthocenter), angle bisectors (incenter), and perpendicular bisectors (circumcenter).
Can I use this for any three cevians?
Yes, Ceva's theorem works for any three cevians, not just the special cases. If arbitrary cevians satisfy the ratio condition, they must be concurrent.
What's the trigonometric form of Ceva's theorem?
There's an equivalent form using sines: sin(∠BAD)/sin(∠DAC) × sin(∠CBE)/sin(∠EBA) × sin(∠ACF)/sin(∠FCB) = 1, where AD, BE, CF are the concurrent cevians.
How is this used in geometry proofs?
Ceva's theorem is essential for proving that certain lines are concurrent, particularly in competition mathematics and when studying triangle centers and their properties.
Is there a 3D version?
Yes! Ceva's theorem has been generalized to tetrahedra and higher dimensions, checking when certain planes or hyperplanes meet at a common point or line.