It is proved that the midpoint of ∵AG is M∴ in Rt△ABG, MB= 12AG.

At Rt△ADG, MD= 12AG.

∴MB=MD.

(2)∫∠BMG =∠BAM+∠ABM = 2∠BAM，

Similarly ∠DMG=∠DAM+∠ADM=2∠DAM,

∴∠BMD=2∠BAM+2∠DAM=2∠BAC，

And < BAC = 90-α,

∴∠BMD= 180 -2α，

∴ When α = 45 and ∠ BMD = 90, then △BMD is an isosceles right triangle.

(3) MB=MD still exists when △CGD rotates counterclockwise around point C for a certain angle.

∠BMD= 180 -2α，

Therefore, when α = 60, △BMD is an equilateral triangle.

Solution: extend DM to n so that MN=DM, even AN, BN, BD, then AN=DH, ∠NAM=∠DHM.

∠∠BAM+90 =∠AHD+90-∠DCB，

∴∠NAB=∠DCB，

∠∠CDH =∠ABC = 90，∠DCH=∠BCA，

∴△CDH∽△CBA，

∴DH:AB=CD:BC，

∴AN:AB=CD:BC，

∴△NAB∽△DCB，

∴∠NBA=∠DBC

∴∠NBD=90，

∴BM=MD，

Get nb: ab = BD: BC from △NAB∽△DCB.

∴△NBD∽△ABC，

∴∠BNM=∠BAC，

∠∠BMD = 2∠BNM

∴∠BMD=2(90 -α)= 180 -2α。