- 初中数学题在矩形ABCD中,对角线AC和BD相交于点G,E为AD
- 在矩形AB中,对角线AC和BD相交于点G,E为AD的中点,连接BE交AC于F,连接FD,若∠BFA=90°。怎么证明:①。△FED和△DEB相似;②。△CFD和△ABG相似?
- (1)
∠BFA=90,∠BAE=90==>AE^2=EF*EB
AE=DE
==>DE^2=EF*EB==>DE/EF=EB/DE
∠FED=∠DEB
==>△FED∽△DEB
(2)
AD∥BC==>AF/CF=AE/BC=AE/AD=1/2==>AF=1/2CF
∠BFA=90,∠ABC=90==>AB^2=AF*AC=AF*2AG=CF*AG
==>AB*CD=CF*AG
==>CF/CD=AB/AG
AB∥CD==>∠DCF=∠BAG
==>△CFD∽△ABG