(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.