题目

完成下面的填空:已知:如图,与互补, , 试说明: . 理由:与互补即 , (已知) , (                  ▲                  ) . (                  ▲                  )又 , (已知) , 即 . (等式的性质)∴                  ▲                  ∥                  ▲                   . (                  ▲                  ) . (两直线平行,内错角相等) 答案:证明:∵∠BAC与∠GCA互补即∠BAC+∠GCA=180°,(已知)∴AB∥DG(同旁内角互补,两直线平行)∴∠BAC=∠ACD.(两直线平行,内错角相等)又∵∠1=∠2,(已知)∴∠BAC-∠1=∠ACD-∠2,即∠EAC=∠FCA.(等式的性质)∴AE∥CF(内错角相等,两直线平行)∴∠E=∠F.(两直线平行,内错角相等)
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