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EF=kEC…1を示す
AE=3/5AB
EC=AC-AE=AC-3/5AB=AB+BC-3/5AB=2/5AB+BC
AF=5/7AB+2/7BC
EF=AF-AE=4/35AB+2/7BC
1に代入すると
4/35AB+2/7BC=2/5 ×kAB+kBC
2/5k=4/35…2
k=2/7…3
3は2を満たすことより証明された。
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