一條幾何(geometry)

廷長 EF 交 BG 於 E" , 則 ㄥBE"F = ㄥDEF = 30° (同位角 , GB//ED) ㄥABG = ㄥBE"F = 30° (內錯角 , AB//EF)y + ㄥAGB + ㄥABG = 180° (△內角和)
y + 70° + 30° = 180°
y = 80°
廷長 GB 交 CD 於 D" , 得 ㄥCD"B = ㄥCDE = 50° (同位角 , GB//ED) ㄥDCB + ㄥCD"B = ㄥCBG (△外角)
40° + 50° = x + ㄥABG
90° = x + 30°
x = 60°

2012-01-12 13:52:43 補充：
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