From ∠BMD=∠BMA+∠AMD=∠C+∠CDM, ∠ B = ∠ AMD = ∠ C = 45, it can be proved that △ ABM ∠△ MCD, and then the similarity is equal to.

∠∠B =∠AMD =∠C = 45，

∴∠BMA=∠CDM，

∴△ABM∽△MCD，

∴ ,

∵M is the midpoint of BC side,

∴MC=BM，

AB = 8，CD=9，

∴BM=MC=6，

∴BC= 12，

Extend the intersection of BA and CD at point e,

∠∠B =∠C = 45，

∴∠E=90，BE=CE，

∴BE=CE= 12，

∴AE=BE-AB=4,DE=CE-CD=3，

At Rt△AED, ad = 5.

So choose C.