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證明:分別過C作CN⊥BE,CH⊥DF,連接CE、CF,∵S△BCE=12S平行四邊形ABCD=S△DFC,∴12•DF•CH=12•BE•CN,∵BE=DF,∴CN=CH,∴GC平分∠BGD(到角兩邊的距離相等的點在角的平分線上).
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