elem.3.34
从给定圆截取一段弓形，使其所含的角等于给定的直线角。
本页以“从圆截取含给定角的弓形”整体图解辅助阅读；点、线、角、圆索引已按命题文字和证明步骤校订，可与证明和问答联动。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
Let ABC be the given circle, and the angle at D the given rectilineal angle; thus it is required to cut off from the circle ABC a segment admitting an angle equal to the given rectilineal angle, the angle at D.
设ABC为给定圆，D为给定直线角。
Let EF be drawn touching ABC at the point B, and on the straight line FB, and at the point B on it, let the angle FBC be constructed equal to the angle at D.
过点B作圆ABC的切线EF，并在直线FB上点B处作角FBC等于角D。
[I. 23] Then, since a straight line EF touches the circle ABC, and BC has been drawn across from the point of contact at B, the angle FBC is equal to the angle constructed in the alternate segment BAC.
由于EF切圆于B，BC为过切点的弦，故角FBC等于相对弓形BAC内的角。
[III. 32] But the angle FBC is equal to the angle at D; therefore the angle in the segment BAC is equal to the angle at D.
角FBC等于角D，因此弓形BAC内的角等于角D。