第3卷命题 3 · 圆心线平分弦则垂直

Let ABC be a circle, and in it let a straight line CD through the centre bisect a straight line AB not through the centre at the point F; I say that it also cuts it at right angles. For let the centre of the circle ABC be taken, and let it be E; let EA, EB be joined. Then, since AF is equal to FB, and FE is common, two sides are equal to two sides; and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE.

设圆ABC，过圆心E的直线CD平分不过圆心的弦AB于点F。连接EA、EB。

[I. 8] But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10] therefore each of the angles AFE, BFE is right. Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles.

因AF等于FB，FE公共，且EA等于EB，故三角形AFE与BFE全等，角AFE等于角BFE。

Again, let CD cut AB at right angles; I say that it also bisects it. that is, that AF is equal to FB.

由定义，相邻两角相等则均为直角，故角AFE和BFE为直角，即CD垂直于AB。

For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5] But the right angle AFE is equal to the right angle BFE, therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore AF is equal to FB.

反之，设CD垂直于AB于F。因EA等于EB，角EAF等于角EBF，且直角AFE等于直角BFE，EF公共，故三角形EAF与EBF全等，AF等于FB。