第3卷命题 17 · 过圆外一点作切线

Let A be the given point, and BCD the given circle; thus it is required to draw from the point A a straight line touching the circle BCD. For let the centre E of the circle be taken; [III. 1] let AE be joined, and with centre E and distance EA let the circle AFG be described; from D let DF be drawn at right angles to EA, and let EF, AB be joined; I say that AB has been drawn from the point A touching the circle BCD.

设A为给定点，BCD为给定圆，取圆心E，连接AE。

For, since E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two sides FE, ED: and they contain a common angle, the angle at E; therefore the base DF is equal to the base AB, and the triangle DEF is equal to the triangle BEA, and the remaining angles to the remaining angles; [I. 4] therefore the angle EDF is equal to the angle EBA.

以E为圆心、EA为半径作圆AFG，从D作DF垂直于EA，连接EF和AB。

But the angle EDF is right; therefore the angle EBA is also right.

由于EA=EF，ED=EB，且角AED为公共角，故三角形DEF全等于三角形BEA，得角EDF等于角EBA。

Now EB is a radius; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16, Por.] therefore AB touches the circle BCD.

角EDF为直角，故角EBA也为直角；EB为半径，过半径外端且垂直于半径的直线与圆相切，因此AB与圆BCD相切。