如图,四边形ABCD是矩形,把矩形沿AC折叠,点B落在点E处,AE与DC的交点为O, 连接DE.(1)求证:?ADE≌?CED;(2)求证: DE∥AC.
(1)证明见解析;(2)证明见解析. |
试题分析:(1)根据矩形的性质和折叠对称的性质,由SSS可证明?ADE≌?CED. (2)根据全等的性质和折叠对称的性质,可求得∠OAC =∠DEA,从而根据平行的判定得出结论. 试题解析:(1)∵ 四边形ABCD是矩形,∴AD=BC,AB=CD. 又∵AC是折痕,∴BC =" CE" =" AD" ,AB =" AE" =" CD" . 又∵DE = ED,∴ΔADE ≌ΔCED(SSS). (2)∵ΔADE ≌ΔCED,∴∠EDC =∠DEA. 又∵ΔACE与ΔACB关于AC所在直线对称,∴∠OAC =∠CAB. 又∵∠OCA =∠CAB,∴∠OAC =∠OCA. ∴2∠OAC = 2∠DEA. ∴∠OAC =∠DEA. ∴DE∥AC. |