IMO 1993 Problem 2

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.)