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Junior high school mathematics cooco
Inspection center: the judgment and nature of similar triangles; Discrimination formula of root; The properties of rectangles.

Answer: (1) Proof: ∵b=2a, and point m is the midpoint of AD.

∴AB=AM=MD=DC=a,

∫ In rectangular ABCD ∠ A = ∠ D = 90,

∴∠AMB=∠DMC=45,

∴∠BMC=90。

(2) solution: existence,

Reason: If BMC = 90,

Then ∠ AMB = ∠ DMC = 90,

And ≈AMB+∠ABM = 90,

∴∠ABM=∠DMC,

∠∠A =∠D = 90,

∴△ABM∽△DMC,

∴=,

Let AM=x, then =,

Ordered: x2-bx+a2 = 0,

∫b > 2a,a>0,b>0,

∴△=b2﹣4a2>0,

The equation has two unequal real roots, both of which are greater than zero, which accords with the meaning of the question.

When b > 2a, there exists BMC = 90,

(3) Solution: Not valid.

Reason: If BMC = 90,

According to (2), x2-bx+a2 = 0,

∫b < 2a,a>0,b>0,

∴△=b2﹣4a2