Solution: crossing n makes NM⊥CN cross CB to m, connecting NA and NB.

∵N is the midpoint of the semi-arc, ∴∠NCB=45 degrees.

So NC=NM

NB=NA, ∠NMB=∠NCA= 135 degrees, ∠NBC=∠NAC.

Therefore ∠MNB=∠CNA.

∴ Triangle MNB≌ Triangle CNA

∴MB=CA

So CB-CA=CB-MB

= cm

=√(CN^2+NM^2)

=√(√3^2+√3^2)

=√6