elem.3.7
在圆的直径上取一个不是圆心的点，从该点向圆引线段，则通过圆心的线段最长，该直径的剩余部分最短，其余线段中，靠近通过圆心的直线的线段总是大于远离的，并且从该点向圆只能引出两条相等的线段，分别位于最短线段的两侧。
本页以“圆内非中心点引线长短定理”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 let AD be a diameter of it; on AD let a point F be taken which is not the centre of the circle, let E be the centre of the circle, and from F let straight lines FB, FC, FG fall upon the circle ABCD; I say that FA is greatest, FD is least, and of the rest FB is greater than FC, and FC than FG. For let BE, CE, GE be joined. Then, since in any triangle two sides are greater than the remaining one, [I. 20] EB, EF are greater than BF. But AE is equal to BE; therefore AF is greater than BF. Again, since BE is equal to CE, and FE is common, the two sides BE, EF are equal to the two sides CE, EF.
连接圆心E与各点，由三角形两边之和大于第三边，得EB+EF>BF，又AE=EB，故AF>BF。
But the angle BEF is also greater than the angle CEF; therefore the base BF is greater than the base CF. [I. 24] For the same reason CF is also greater than FG. Again, since GF, FE are greater than EG, and EG is equal to ED, GF, FE are greater than ED. Let EF be subtracted from each; therefore the remainder GF is greater than the remainder FD.
在三角形BEF和CEF中，BE=CE，EF公共，角BEF>角CEF，由边角边定理得BF>CF。同理CF>FG。
Therefore FA is greatest, FD is least, and FB is greater than FC, and FC than FG. I say also that from the point F only two equal straight lines will fall on the circle ABCD, one on each side of the least FD. For on the straight line EF, and at the point E on it, let the angle FEH be constructed equal to the angle GEF [I. 23], and let FH be joined. Then, since GE is equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the angle GEF is equal to the angle HEF; therefore the base FG is equal to the base FH.
由GF+FE>EG，EG=ED，得GF+FE>ED，减去EF得GF>FD，故FA最大，FD最小。
[I. 4] I say again that another straight line equal to FG will not fall on the circle from the point F. For, if possible, let FK so fall. Then, since FK is equal to FG, and FH to FG, FK is also equal to FH, the nearer to the straight line through the centre being thus equal to the more remote: which is impossible. Therefore another straight line equal to GF will not fall from the point F upon the circle; therefore only one straight line will so fall.
作角FEH等于角GEF，连接FH，由全等三角形得FG=FH。若另有FK=FG，则FK=FH，与靠近中心线者大于远离者矛盾，故只有两条相等线段。