
已知:平行四边形ABCD中,点E是AB的中点,在直线AD上截取AF=2FD,EF交AC于G,则AGAC=27或2527或25
已知:平行四边形ABCD中,点E是AB的中点,在直线AD上截取AF=2FD,EF交AC于G,则AGAC=27或2527或25....
已知:平行四边形ABCD中,点E是AB的中点,在直线AD上截取AF=2FD,EF交AC于G,则AGAC=27或2527或25.
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解答:
解:(1)点F在线段AD上时,设EF与CD的延长线交于H,
∵AB∥CD,
∴△EAF∽△HDF,
∴HD:AE=DF:AF=1:2,
即HD=
AE,
∵AB∥CD,
∴△CHG∽△AEG,
∴AG:CG=AE:CH
∵AB=CD=2AE,
∴CH=CD+DH=2AE+
AE=
AE,
∴AG:CG=2:5,
∴AG:(AG+CG)=2:(2+5),
即AG:AC=2:7;
(2)点F在线段AD的延长线上时,设EF与CD交于H,
∵AB∥CD,
∴△EAF∽△HDF,
∴HD:AE=DF:AF=1:2,
即HD=
AE,
∵AB∥CD,
∴△CHG∽△AEG,
∴AG:CG=AE:CH
∵AB=CD=2AE,
∴CH=CD-DH=2AE-
AE=
AE,
∴AG:CG=2:3,
∴AG:(AG+CG)=2:(2+3),
即AG:AC=2:5.
故答案为:
或
.
∵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)点F在线段AD的延长线上时,设EF与CD交于H,
∵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.
故答案为:
2 |
5 |
2 |
7 |
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