IMO 2004 SL G4

Origin: POL | Category: Geometry

Problem

In a convex quadrilateral ABCD the diagonal BD does not bisect the angles ABC and CDA. The point P lies inside ABCD and satisfies \anglePBC = \angleDBA and \anglePDC = \angleBDA. Prove that ABCD is a cyclic quadrilateral if and only if AP = CP.

Solution

Assume first that the points A, B, C, D are concyclic. Let the lines BP and DP meet the circumcircle of ABCD again at E and F, respectively. Then it follows from the given conditions that : AB = : CF and : AD = : CE; hence BF \parallelAC and DE \parallelAC. Therefore BFED and BFAC are isosceles trapezoids and thus P = BE\capDF lies on the common bisector of segments BF, ED, AC. Hence AP = CP. Assume in turn that AP = CP. Let P w.l.o.g. lie in the triangles ACD and BCD. Let BP and DP meet AC at K and L, respectively. The points A and C are isogonal conjugates with respect to \triangleBDP, which implies that \angleAPK = \angleCPL. Since AP = CP, we infer that K and L are symmetric with respect to the perpendicular bisector p of AC. Let E be the reflection of D in p. Then E lies on the line BP, and the triangles APD and CPE are congruent. Thus \angleBDC = \angleADP = \angleBEC, which means that the points B, C, E, D are concyclic. Moreover, A, C, E, D are also concyclic. Hence, ABCD is a cyclic quadrilateral.