（6分）如圖，在△ABC中，AB=AC， AD⊥BC，垂足為D，AE∥BC, DE∥AB.
證明：（1）AE=DC；
（2）四邊形ADCE為矩形．
證明：
（1）在△ABC中，∵AB=AC，AD⊥BC，
∴BD=DC······························································································ 1分
∵AE∥BC, DE∥AB，
∴四邊形ABDE為平行四邊形······································································ 2分
∴BD=AE，···························································································· 3分
∵BD=DC
∴AE = DC．·························································································· 4分
（2）
解法一：∵AE∥BC，AE= DC，
∴四邊形ADCE為平行四邊形．··································································· 5分
又∵AD⊥BC，
∴∠ADC=90°，
∴四邊形ADCE為矩形．··········································································· 6分
解法二：
∵AE∥BC，AE= DC，
∴四邊形ADCE為平行四邊形······································································ 5分
又∵四邊形ABDE為平行四邊形
∴AB=DE．∵AB=AC，∴DE=AC．
∴四邊形ADCE為矩形．··········································································· 6分
解析:略
閱讀快車系列答案科目：初中數學 來源： 題型：
如圖，在△ABC中，∠ACB=90°，AC=BC=1，取斜邊的中點，向斜邊作垂線，畫出一個新的等腰三角形，如此繼續下去，直到所畫出的直角三角形的斜邊與△ABC的BC重疊，這時這個三角形的斜邊為A、
| ||||
B、（
| ||||
C、
| ||||
D、
|
查看答案和解析>>
湖北省互聯網違法和不良信息舉報平臺 | 網上有害信息舉報專區 | 電信詐騙舉報專區 | 涉歷史虛無主義有害信息舉報專區 | 涉企侵權舉報專區
違法和不良信息舉報電話：027-86699610 舉報郵箱：58377363@163.com
20、如圖，在△ABC中，∠BAC=45°，現將△ABC繞點A逆時針旋轉30°至△ADE的位置，使AC⊥DE，則∠B=
2、如圖，在△ABC中，DE∥BC，那么圖中與∠1相等的角是（ ）
如圖，在△ABC中，AB=AC，且∠A=100°，∠B=