※ 引述《blackymys (mys)》之銘言：
: https://i.mopix.cc/WtzFus.jpg
: 請教各位先進，此題怎麼證明，感謝！
設AG交BD於M，令△AFG = 2a，△AEG = 2b，△GBC = 2c
=> △FMC = c(1 - GF/GD)，△GFC = (2/3)c(1 - GF/GD)
類似可得
△EBM = c(1 + GE/GD)，△EGB = (2/3)c(1 + GE/GD)
=> a = c + (2/3)c(1 - GF/GD) - c(1 - GF/GD) = (c/3)[2 + GF/GD]
同理b = (c/3)[2 - GE/GD]
=> GF/GE = [2 + GF/GD]/[2 - GE/GD]
=> 2GF/GE - GF/GD = 2 + GF/GD
=> 1/GE = 1/GF + 1/GD Q.E.D.
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