Menelaus' Theorem

fraction of the distance of F from A to C

0.356

AD	≈	2	BE	≈	0.32	CF	≈	0.96
BD	≈	1	CE	≈	1.16	AF	≈	0.53

AD

×

BE

×

CF

≈

0.62

BD

×

CE

×

AF

≈

0.62

In triangle ABC, let a line DEF intersect the two sides AC and BC and an extension of AB at F, E, and D. Then AD

×

BE

×

CF = BD

×

CE

×

AF.

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Jay Warendorff