![](http://hiphotos.baidu.com/zhidao/pic/item/8694a4c27d1ed21b86e65d59a86eddc451da3f5a.jpg)
∵AB∥CD,
∴△EAF∽△HDF,
∴HD:AE=DF:AF=1:2,
即HD=
1 |
2 |
∵AB∥CD,
∴△CHG∽△AEG,
∴AG:CG=AE:CH
∵AB=CD=2AE,
∴CH=CD+DH=2AE+
1 |
2 |
5 |
2 |
∴AG:CG=2:5,
∴AG:(AG+CG)=2:(2+5),
即AG:AC=2:7;
(2)點(diǎn)F在線段AD的延長(zhǎng)線上時(shí),設(shè)EF與CD交于H,
![](http://hiphotos.baidu.com/zhidao/pic/item/f9dcd100baa1cd1125262cdcba12c8fcc2ce2dbd.jpg)
∵AB∥CD,
∴△EAF∽△HDF,
∴HD:AE=DF:AF=1:2,
即HD=
1 |
2 |
∵AB∥CD,
∴△CHG∽△AEG,
∴AG:CG=AE:CH
∵AB=CD=2AE,
∴CH=CD-DH=2AE-
1 |
2 |
3 |
2 |
∴AG:CG=2:3,
∴AG:(AG+CG)=2:(2+3),
即AG:AC=2:5.
故答案為:
2 |
5 |
2 |
7 |