Consider the convex quadrilateral ABCD. The point P is in the interior of ABCD. The following ratio equalities hod:
∠PAD:∠PBA:∠DPA=1:2:3=∠CBP:∠BAP:∠BPC.
Prove that the following three lines meet in a point : the internal bisectors of angles ∠ADP and ∠PCB and the perpendicular bisector of segment AB.
设P是凸四边形ABCD内部一点,且满足:
∠PAD:∠PBA:∠DPA=1:2:3=∠CBP:∠BAP:∠BPC.
证明:∠ADP的内角平分线、∠PCB的内角平分线和线段AB的中垂线,三线共点。
(波兰供题)