[alogo] Four tangent circles

Given a convex cyclic quadrilateral q = ABCD, inscribed in a circle (c), there is exactly one point P on its plane such that the two pairs of circles {PAB}, {PCD} and {PBC}, {PAD} are pairs of exterior tangent circles. P is one intersection point of the orthogonal from the circumcenter to the diagonal EF of q with the circle (d) having diameter EF. The common tangents pass through E and F correspondingly.

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Notice that line OP contains the intersection point G of the diagonals AC and BD. G is the pole of EF with respect to c. Circles c and d are orthogonal and G is on the radical axis of c and d. For every point of the plane the corresponding pairs of circles {PAB}, {PCD} and {PBC}, {PAD} have radical axes passing through F and E correspondingly. Thus, orthogonality of the radical axes is possible only when P is on the circle (d). For points P' on circle (d) tangency of the corresponding pairs occurs only on P and Q. Exterior tangency occurs only at P. The centers of the corresponding circles form an orthodiagonal quadrilateral q' = HIJK. Its "double", q* = LMNO, homothetic of q' w.r. to P at ratio 2, is circumscribed to the original quadrangle q. There are lots of interesting properties in this figure. Some of them are discussed further in the file OrthodiagonalFromCyclic.html . Another, to be proved, property is that the opposite sides of q' intersect on line EF. Finally the study of the other four circles, passing through Q deserves a discussion.

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