已知菱形ABCD,AB=4,∠ADC=120°,点M在射线AB上,BM=1,∠DMN=60°,射线MN交射线BC于N,则BN=______

①当点M在线段AB上时,

过点D作DQ⊥AB于点Q,连接BD,延长DM、CB交于一点P,

则AQ=BQ=2,QM=BM=1,DQ=2

3

在Rt△DQM中,DM=

(2
3
)2+12
=
13

∵BC∥AD,

BP
AD
=
PM
DM
=
BM
AM
=
1
3

解得:BP=

4
3
,PM=
13
3

∵∠DMN=60°,∠DBC=60°,

∴∠PMN=120°,∠PBD=120°,

∴△PMN∽△PBD,

PN
PD
=
PM
PB
,即
4
3
+BN
13
+
13
3
=
13
3
4
3

解得:BN=3;

②当点M在AB延长线上时,

过点D作DQ⊥AB于点Q,连接BD,

则DQ=2

3
,BD=4,DM=
DQ2+QM2
=
21

∵BP∥AD,

MP
PD
=
BM
AB
=
1
4
BP
AD
=
MB
AM
=
1
5

∴BP=

4
5

又∵MP+PD=DM=

21

∴MP=

21
5
,PD=
4
21
5

∵∠PMN=∠DMN=60°,∠PBD=∠CBD=60°,

∴△PBD∽△PMN,

PM
PB
=
PN
PD
,即
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