Proof of The Theorem
A standard proof is as follows:
First, the sign of the left-hand side is positive since either all three of the ratios are positive, the case where O is inside the triangle (upper diagram), or one is positive and the other two are negative, the case O is outside the triangle (lower diagram shows one case).
To check the magnitude, note that the area of a triangle of a given height is proportional to its base. So
Therefore,
(Replace the minus with a plus if A and O are on opposite sides of BC.) Similarly,
and
Multiplying these three equations gives
as required.
The theorem can also be proven easily using Menelaus' theorem. From the transversal BOE of triangle ACF,
and from the transversal AOD of triangle BCF,
The theorem follows by dividing these two equations.
The converse follows as a corollary. Let D, E and F be given on the lines BC, AC and AB so that the equation holds. Let AD and BE meet at O and let F′ be the point where CO crosses AB. Then by the theorem, the equation also holds for D, E and F′. Comparing the two,
But at most one point can cut a segment in a given ratio so F=F′.
Read more about this topic: Ceva's Theorem
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