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When the quadrilateral AEDF is a square,

AE=DE

Because DE=BE

vertical

So the triangle DBE is an isosceles right triangle.

Angle B= angle c = 45.

So the triangle ABC is an isosceles right triangle.

Let AB intersect with the vertical straight line and point d.

Then OD=AOcos60=BOcos30.

AO=V a *t

BO=V b *t

Solvable v b

I don't know how to apply for adoption. thank you

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