Theorem of Menelaus

Line d intersects the sides of triangle ABC in $ E\in AC,F\in AB,D\in BC $.
Then

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


Proof
Let us consider G the point in which parallel by B at AC intersects EF.


$ \Delta AFE  $ is similar to $ \Delta BFG $

$$\frac{AF}{FB}=\frac{AE}{BG}$$

Also $ \Delta DBG $ similar to $ \Delta DCE $

$$\frac{BD}{DC}=\frac{BG}{CE}$$

Then we have

$$\frac{AF}{FB}\frac{BD}{DC}=\frac{AE}{BG}\frac{BG}{CE}=\frac{EA}{CE}$$

which can be written

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