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WOLFRAM|DEMONSTRATIONS PROJECT

Fuhrmann's Theorem

​
AB
≈
0.85
ED
≈
0.69
BC
≈
1.21
EF
≈
0.84
CD
≈
1.54
FA
≈
0.77
AD
≈
1.85
BE
≈
1.91
CF
≈
1.99
AD
×
BE
×
CF
≈
7.02
AB
×
ED
×
CF
+
AF
×
CD
×
BE
+
BC
×
EF
×
AD
+
AB
×
CD
×
EF
+
DE
×
FA
×
BC
≈
7.02
Let AB and ED, BC and EF, and CD and FA be three pairs of opposite sides in a convex cyclic hexagon ABCDEF. Then
AD×BE×CF=AB×ED×CF+AF×CD×BE+BC×EF×AD+AB×CD×EF+DE×FA×BC
.
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