elem.3.37
若圆外一点到圆上作两线段，其中一条割圆，另一条落于圆上，且割线全长与圆外部分所成矩形等于落于圆上线段上的正方形，则落于圆上的线段切于该圆。
本页以“圆外点割线等积证切线”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。
For let a point D be taken outside the circle ABC; from D let the two straight lines DCA, DB fall on the circle ACB; let DCA cut the circle and DB fall on it; and let the rectangle AD, DC be equal to the square on DB. I say that DB touches the circle ABC. For let DE be drawn touching ABC; let the centre of the circle ABC be taken, and let it be F; let FE, FB, FD be joined.
设圆ABC外一点D，作线段DCA割圆，DB落于圆上，且矩形AD·DC等于DB上的正方形。
Thus the angle FED is right. [III. 18] Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE.
作DE切圆于E，取圆心F，连接FE、FB、FD，则∠FED为直角。
[III. 36] But the rectangle AD, DC was also equal to the square on DB; therefore the square on DE is equal to the square on DB; therefore DE is equal to DB. And FE is equal to FB; therefore the two sides DE, EF are equal to the two sides DB, BF; and FD is the common base of the triangles; therefore the angle DEF is equal to the angle DBF. [I. 8] But the angle DEF is right; therefore the angle DBF is also right.
由切线性质，矩形AD·DC等于DE上的正方形，又等于DB上的正方形，故DE等于DB。
And FB produced is a diameter; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16, Por.] therefore DB touches the circle. Similarly this can be proved to be the case even if the centre be on AC.
因FE等于FB，FD公共，三角形DEF与DBF全等，得∠DBF为直角，故DB切于圆。