Call the point where BK intersects AM point X. BX is the altitude of ∆ABM, and it is also the angle bisector of angle ∠ABM. Hence ∆ABM is isosceles, and AB ≅ MB.
M is the midpoint of BC, which is 12 inches long. Hence MB is half that length, or 6 inches.
Since AB is the same length as MB, which is 6 inches, ...
... AB is 6 inches.
