Analysis: (1) According to the nature of the midline on the hypotenuse of the right triangle, BM=DM, then ∠BME=2∠BCM, ∠EMD=2∠DCM, then ∠ BMD = 90 According to the nature of the isosceles right triangle,

(2) Extend the intersection of DM and BC at point n, and prove △ EDM △ CNM, and deduce AD=CN, BD=BN, BM= DN=DM, and then deduce BM⊥DN, and then "△BMD is an isosceles right triangle".

Answer: (1) Proof: As shown in the figure,

∠△ABC and △ADE are isosceles right triangles, ∠ ABC = ∠ ade = 90,

∴∠EDC=90, ba = BC,

∴∠BCA=45，

Point m is the midpoint of EC,

∴bm=( 1/2)ec=mc,dm=( 1/2)EC = MC，

∴BM=DM，

∴∠MBC=∠MCB,∠MDC=∠MCD，

∴∠BME=2∠BCM,∠EMD=2∠DCM，

∴∠BMD=∠BME+∠EMD=2∠BCM+2∠DCM

=2(∠BCM+∠DCM)=2∠BCA=2×45 =90，

∴△BMD is an isosceles right triangle.

(2) Solution: △BMD is an isosceles right triangle for the following reasons:

Extend DM to point n of BC.

∠△ABC and △ADE are isosceles right triangles, ∠ ABC = ∠ ade = 90,

∴BA=BC,DE=DA,∠EDB=90，

∴∠EDB=∠DBC，

∴ED∥BC，

∴∠DEC=∠BCE，

Point m is the midpoint of EC,

∴EM=CM，

∵ In △EDM and △CNM, ∠DEM=∠NCM, EM=CM, ∠EMD=∠CMN,

∴△EDM≌△CNM，

∴ED=CN,MD=MN，

∴AD=CN，

∴BA-DA=BC-NC，

That is, BD=BN,

∴BM= ( 1/2)DN=DM，

∴BM⊥DN, that is ∠ BMD = 90,

∴△BMD is an isosceles right triangle.