Ceva's theorem

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Ceva's theorem, case 1: the three lines are concurrent at a point O inside ABC
Ceva's theorem, case 1: the three lines are concurrent at a point O inside ABC
Ceva's theorem, case 2: the three lines are concurrent at a point O outside ABC
Ceva's theorem, case 2: the three lines are concurrent at a point O outside ABC

Ceva's theorem is a well-known theorem in elementary geometry. Given a triangle ABC, and points D, E, and F that lie on lines BC, CA, and AB respectively, the theorem states that lines AD, BE and CF are concurrent if and only if

\frac{AF}{FB}  \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.

There is also an equivalent trigonometric form of Ceva's Theorem, that is, AD,BE,CF concur if and only if

\frac{\sin\angle BAD}{\sin\angle CAD}\times\frac{\sin\angle ACF}{\sin\angle BCF}\times\frac{\sin\angle CBE}{\sin\angle ABE}=1.

The theorem was proved by Giovanni Ceva in his 1678 work De lineis rectis, but it was also proved much earlier by Al-Mu'taman ibn Hűd, an eleventh-century king of Saragossa.

Associated with the figures are several terms derived from Ceva's name: cevian (the lines AD, BE, CF are the cevians of O), cevian triangle (the triangle DEF is the cevian triangle of O); cevian nest, anticevian triangle, Ceva conjugate. (Ceva is pronounced Chay'va; cevian is pronounced chev'ian.)

Contents

Suppose AD, BE and CF intersect at a point O. Because \triangle BOD and \triangle COD have the same height, we have

\frac{|\triangle BOD|}{|\triangle COD|}=\frac{BD}{DC}.

Similarly,

\frac{|\triangle BAD|}{|\triangle CAD|}=\frac{BD}{DC}.

From this it follows that

\frac{BD}{DC}= \frac{|\triangle BAD|-|\triangle BOD|}{|\triangle CAD|-|\triangle COD|} =\frac{|\triangle ABO|}{|\triangle CAO|}.

Similarly,

\frac{CE}{EA}=\frac{|\triangle BCO|}{|\triangle ABO|},

and

\frac{AF}{FB}=\frac{|\triangle CAO|}{|\triangle BCO|}.

Multiplying these three equations gives

\frac{AF}{FB}  \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1,

as required. Conversely, suppose that the points D, E and F satisfy the above equality. Let AD and BE intersect at O, and let CO intersect AB at F'. By the direction we have just proven,

\frac{AF '}{F 'B}  \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.

Comparing with the above equality, we obtain

\frac{AF '}{F 'B}=\frac{AF}{FB}.

Adding 1 to both sides and using AF' + F'B = AF + FB = AB, we obtain

\frac{AB}{F 'B}=\frac{AB}{FB}.

Thus F'B = FB, so that F and F' coincide (recalling that the distances are directed). Therefore AD, BE and CF=CF' intersect at O, and both implications are proven.

  • J. B. Hogendijk, "Al-Mutaman ibn Hűd, 11the century kin of Saragossa and brilliant mathematician," Historia Mathematica 22 (1995) 1-18.
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