Math Magazine Problem 1573 Solution.

    Problem. (by Jiro Fukuta, Cifu Univ., Japan) AD and BDC are two distinct lines intersecting at D. E is a point on line AD. Let line BE intersect circle ACD at M, N, and line CE intersect circle ABD at P,Q. Prove that M,N,P,Q lie on some circle whose center lies on a line through D perpendicular to BDC.

    Proof. ME × EN = AE × ED (because M, N, A, D are concyclic) = PE × EQ (because P, Q, A, D are concyclic). Hence M, N, P, Q are concyclic.
    It remains to prove that the center Z of circle MPNQ satisfies ZD ^ BC. Let O and O' be the circumcenters of circles ADB and ADC. Let K, K' be the midpoints of the chords MN and PQ, and H, H' be the midpoints of BD and DC. In Diagram 2 we drop off the circles, the lines AB, AC, and the points M, N, P, Q, because we know the lines OK and O'K' intersect at Z. We need only to prove ZD ^ BC.

     Let JDT' be a line parallel to OKZ intersecting lines EC, OH at T' and J. Let J'DT be a line parallel to O'K'Z intersecting lines EB, O'H' at T and J'. We shall prove triangles ZOO' and DJJ' are congruent.
     First, since B,T,H',J' are concyclic, ÐDJH' = ÐB, so that DJ' = DH' / sin B = (½) DC / sin B = (½) DT' / sin C sin B. Similarly, DJ = (½) DT / sin B sin C. Thus DJ' / DT' = DJ / DT. Hence T,T',J',J are concyclic. Hence Ð DJ'J = ÐDT'T = ÐDET because D,T,E,T' are concyclic. Since TE ^ DJ', hence ED ^ JJ'. Now OO' is the perpendicular bisector of AD, hence OO' is parallel to JJ'. Therefore triangles ZOO', and DJJ' are congruent as well as having pairwise parallel sides. Hence ZDJO is a parallelogram, and ZD is then perpendicular to BC. QED

Remarks.The above proof avoided trigonometrical manipulations, and is more purely geometrical.

     If A,B,C,D are fixed points, and E is a variable point moving along the infinite line AD, then the set of circles PQMN associated with each position of E form a coaxial family of circles whose centers lie on the line through D perpendicular to BC.