Theorem of Menelaus

Posted January 1st, 2008 by Structure

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

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

Instant start