试题(2020年9月国际数学奥林匹克

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)

参考答案

关键词

数学;平面几何;三角形;四边形;证明;命题;平分线;中垂线;adp;