IMO Shortlist 1993 problem G7


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2. travnja 2012.
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Let A, B, C, D be four points in the plane, with C and D on the same side of the line AB, such that AC \cdot BD = AD \cdot BC and \angle ADB = 90^{\circ}+\angle ACB. Find the ratio
\frac{AB \cdot CD}{AC \cdot BD},
and prove that the circumcircles of the triangles ACD and BCD are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles ACD and BCD are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles ACD and BCD at the point C are perpendicular.)
Izvor: Međunarodna matematička olimpijada, shortlist 1993