题目内容
如图,已知在平行四边形ABCD中,点E、F分别在边AB、CD上,且AE=2EB,CF=2FD,连接EF.
(1)写出与
相等的向量______;
(2)填空
+
-
=______;
(3)求作:
-
.(保留作图痕迹,不要求写作法,请说明哪个向量是所求作的向量)

(1)写出与
FC |
(2)填空
AD |
EB |
EF |
(3)求作:
AD |
FE |

(1)在?ABCD中,AB∥CD,AB=CD,
∵AE=2EB,CF=2FD,
∴AE=
AB=
AB,
CF=
CD=
CD,
∴与
相等的向量是
;
(2)如图,连接AF,
∵DF=CD-FC=
CD,
BE=AB-AE=
AB,
∴
=
,
∴
+
=
,
∵-
=
,
∴
-
=
+
=
,
又∵
=
,
∴
+
-
=
(或
);
故答案为:(1)
;(2)
或
;
(3)如图,
即为所求作的
-
.
∵AE=2EB,CF=2FD,
∴AE=
2 |
1+2 |
2 |
3 |
CF=
2 |
1+2 |
2 |
3 |
∴与
FC |
AE |

(2)如图,连接AF,
∵DF=CD-FC=
1 |
3 |
BE=AB-AE=
1 |
3 |
∴
EB |
DF |

∴
AD |
EB |
AF |
∵-
EF |
FE |
∴
AF |
EF |
AF |
FE |
AE |
又∵
AE |
FC |
∴
AD |
EB |
EF |
AE |
FC |
故答案为:(1)
AE |
AE |
FC |
(3)如图,
GD |
AD |
FE |

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