急 (17 21:14:44)如图四边形ABCD是矩形(AD>AB),点E在BC上,且AE=AD,DF垂直于AE,垂足为F,求证:AB=DFA                      

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/16 14:59:50
急 (17 21:14:44)如图四边形ABCD是矩形(AD>AB),点E在BC上,且AE=AD,DF垂直于AE,垂足为F,求证:AB=DFA                      
xTJ@Ab(hRlC?@@ m^0^塍^_bv nwD9=3g3S

急 (17 21:14:44)如图四边形ABCD是矩形(AD>AB),点E在BC上,且AE=AD,DF垂直于AE,垂足为F,求证:AB=DFA                      
急 (17 21:14:44)
如图四边形ABCD是矩形(AD>AB),点E在BC上,且AE=AD,DF垂直于AE,垂足为F,求证:AB=DF
A                                 D
 
                     F
B                          E      C   (连接起来即是

急 (17 21:14:44)如图四边形ABCD是矩形(AD>AB),点E在BC上,且AE=AD,DF垂直于AE,垂足为F,求证:AB=DFA                      
∵AE=AD
∴ ∠AED=∠ADE
∵矩形ABCD
∴AD//BC
∴∠ADE=∠DEC
∴∠AED=∠DEC
∵DF⊥AE DC⊥BC DE=DE
⊿DFE≌⊿DCE
∴CD=DF
∴AB=DF

连结DE,∵△AED的面积=1/2 * AD * AB=1/2 * AE * DF(其实就是等积法)
又∵AE=AD
∴DF=AB

同意柯藤的回答,回答得比较具体,思路正确。