elem.6.10
将一条给定的未分线段与一条给定的已分线段按相同比例分割。
把直线 AB 按已分线 AC（在 D、E 处分割）的比例分割：连 BC，过 D、E 作平行于 BC 的辅助线，交 AB 于 F、G；H、K 为构造过程中产生的辅助交点。
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 AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31] Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB.
设AB为未分线段，AC为已分线段，点D、E在AC上。放置它们使成任意角，连接CB。
[I. 34] Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD.
过D、E作DF、EG平行于BC，过D作DHK平行于AB。则FH、HB为平行四边形，故DH等于FG，HK等于GB。
[VI. 2] But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF. Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA.
在三角形DKC中，HE平行于KC，因此CE比ED等于KH比HD。但KH等于BG，HD等于GF，故CE比ED等于BG比GF。
[VI. 2] But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA.
在三角形AGE中，FD平行于GE，因此ED比DA等于GF比FA。结合前一步，得CE比ED等于BG比GF，且ED比DA等于GF比FA。