连接AM,
∵AB=AC,点M为BC中点,
∴AM⊥CM（三线合一）,BM=CM,
∵AB=AC=5,BC=6,
∴BM=CM=3,
在Rt△ABM中,AB=5,BM=3,
∴根据勾股定理得：AM=
AB2−BM2
=
52−33
=4,
又S△AMC=
1
2
MN•AC=
1
2
AM•MC,
∴MN=
AM•CM
AC
=
12
5