延长CM交AB于D,
∵AM⊥CM,∴∠AMC=∠AMD,
∵AM平分∠BAC,∴∠MAC=∠MAD,
∵AM=AM,
∴ΔAMC≌ΔAMD,
∴DM=CM,AD=AC=3,
M为CD中点,
∵N为BC中点,
∴MN=1/2BD=1/2(AB-AD)=1.
延长CM交AB于D,
∵AM⊥CM,∴∠AMC=∠AMD,
∵AM平分∠BAC,∴∠MAC=∠MAD,
∵AM=AM,
∴ΔAMC≌ΔAMD,
∴DM=CM,AD=AC=3,
M为CD中点,
∵N为BC中点,
∴MN=1/2BD=1/2(AB-AD)=1.