The applet is supposed to suggest the folloing proposition:

Denote the centers of the circles A₁, ..., A₄ and the corresponding vertices of the quadrilateral in question B₁, ..., B₄. Since the four circles have the same radius, the four points A₁, A₂, A₃, A₄ are concyclic and lie on a circle with center at the common point of the four given circles and radius equal to their common radius. Thus the quadrilateral A₁A₂A₃A₄ is cyclic. This in particular means that the opposite angles in A₁A₂A₃A₄ add up to 180°.

Turning to the quadrilateral B₁B₂B₃B₄, note that its sides are parallel to the corresponding sides of A₁A₂A₃A₄. For example, A₁A₂||B₁B₂. This is because the points A₁ and A₂ lie at the same distance from the common tangent B₁B₂ to the two circles. The corresponding angles of the two quadrilaterals are, therefore, equal. In particular, this implies that the opposite angles in B₁B₂B₃B₄ add up to 180°. The quadrilateral B₁B₂B₃B₄ is therefore cyclic.

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