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Given a triangle $ABC$, and points $D$, $E$, and $F$ on the sides $BC$, $CA$, and $AB$ respectively, then the segments $AD$, $BE$ and $CF$ are concurrent if and only if

$$\dfrac{AF}{FB}\dfrac{BD}{DC}\dfrac{CE}{EA}=1$$

Suppose that the three segments are concurrent in a point $O$



Then the ratio of lengths $\frac{AF}{FB}$ is equal to the ratio of areas $\frac{AFC}{FBC}$ and $\frac{AFO}{FBO}$, for having equal altitudes. Therefore, following the same proportion, it is also true that $$\dfrac{AF}{FB}=\dfrac{OCA}{OBC}$$ Applying the same arguments to the other sides, it turns out that $$\dfrac{AF}{FB}\dfrac{BD}{DC}\dfrac{CE}{EA}=\dfrac{OCA}{OBC}\dfrac{OAB}{OCA}\dfrac{OBC}{OCA}=1$$