如图,在三角形abc中,角acb等于90度,ac等于bc,过b,c两点的圆o交ab于点p，点e是

郭敦顒回答：
未见图。
∵在△ABC中，∠ACB=90°，AC=BC，过B,C两点的圆O交AB于点P，点E是PB上一点，点D在CA的延长线上，且P为三角形CDE的内心，

∴应有∠ACP＞45°，点E是PB延长线上一点，以此作答——
确定点P后其下的作图步骤是：
（1）作∠PCE=∠ACP交AB延长线于E，
（2）作∠AED=∠AEC交CA延长线于D。
∵点P在∠DCE与∠CED的角平分线上，
∴点P是△CDE的内心。
给出AB与O圆半径OA（OB）的具体值可求出相关线段和角度值。

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