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几何部分rn1.已知△ABC为等腰三角形,BC=CA,D为边AB上的点,使得AD＜DB,点P、Q分别为边BC、CA上的点,使得∠DPB=∠DQA=90°,PQ的中垂线与线段CQ交于点E,△ABC的外接圆与△CPQ的外接圆的第二个交点为F.若P、E、F三点共线,证明:rn∠A CB=90°.
第61届IMO预选题(三)
【摘 要】
：
几何部分rn1.已知△ABC为等腰三角形,BC=CA,D为边AB上的点,使得AD＜DB,点P、Q分别为边BC、CA上的点,使得∠DPB=∠DQA=90°,PQ的中垂线与线段CQ交于点E,△ABC的外接圆与△CPQ的外接圆的第二个交点为F.若P、E、F三点共线,证明:rn∠A CB=90°.
【出 处】
：
中等数学
【发表日期】
：
2021年11期
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