WOLFRAM|DEMONSTRATIONS PROJECT

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.
Drag the slider or the orange points to change the figure.