elem.3.4
在圆内，如果两条不通过圆心的直线互相相交，则它们不会互相平分。
本页以“圆内非直径弦互不相分”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 ABCD be a circle, and in it let the two straight lines AC, BD, which are not through the centre, cut one another at E; I say that they do not bisect one another. For, if possible, let them bisect one another, so that AE is equal to EC, and BE to ED; let the centre of the circle ABCD be taken [III. 1], and let it be F; let FE be joined.
设圆ABCD内，两条不通过圆心的直线AC、BD相交于E。假设它们互相平分，即AE等于EC，BE等于ED。
Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [III. 3] therefore the angle FEA is right.
取圆心F，连接FE。由于过圆心的直线FE平分不通过圆心的弦AC，根据III.3，它也垂直于AC，因此角FEA为直角。
Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles; [III. 3] therefore the angle FEB is right. But the angle FEA was also proved right; therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible.
同理，FE平分弦BD，故它也垂直于BD，因此角FEB也为直角。
Therefore AC, BD do not bisect one another.
但角FEA已被证明为直角，所以角FEA等于角FEB，小角等于大角，不可能。因此AC、BD不能互相平分。