如图,在正方形ABCD中,E为BC边上一点,AF⊥AE交CD延长线于点F,AG平分∠FAE交CD于点G,连接EG,求证:

(1)BE=DF; (2)BE+DG=EG
2025-01-01 01:57:31
推荐回答(2个)
回答1:

证明:(1)∵∠BAE=∠DAF,AB=AD,∠B=∠ADF=90°,
∴△ABE≌△ADF,
∴AE=AF,BE=DF.
(2)∵AG为∠EAF的角平分线,
∴∠EAG=∠FAG,
又∵AE=AF,
∴△AEG≌△AFG,
∴EG=FG,
∵FG=DG+FD,
∴EG=BE+DG.

回答2:

AD=AB
ADF,ABE全等
BE=DF

AGF,AGE全等
EG=DG+FD=DG+BE