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