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。