Question

已知四边形ABCD,点E是BC上一点,角AED=90度,∠BAE=∠DAE,∠ADE=∠CDE,求AB+DC=AD

Answer 1

在AD上取一点F,使得AF＝AB,连接点FE
∵AB＝AF
∠BAE=∠FAE
AE=AE
∴△BAE≌△FAE
∠FEA=∠BEA
又∠FEA+∠DEF=90°
∠BEA+∠CED=90°
则∠FED=∠CED
且DE=DE
∠ADE=∠CDE
∴△FDE≌△CDE
则CD=DF
又AB=AF
∴AB+CD=DF+AF=AD