※ 引述《revengeiori (大笨宗)》之銘言：
: 請教一下版上前輩們這題
: http://i.imgur.com/aNe5UCE.jpg
: 先謝過了QQ
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: Sent from JPTT on my Sony H9493.
延長BH交GC於H'
GD/GH' * HH'/HA = 1 => GH'/HH' = GD/HA = (GC - CD)/(HB - AB)
GC/GH' * HH'/HB = 1 => GH'/HH' = GC/HB
=> HB = GC
過G做HB平行線交BC於B'
GB' : B'E = HB : EB = GC : EC
=> ∠1 = ∠2
原命題得證
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