第1卷命题 2 · 从给定点作一线等于给定直线

Let A be the given point, and BC the given straight line. Thus it is required to place at the point A (as an extremity) a straight line equal to the given straight line BC. From the point A to the point B let the straight line AB be joined; [Post. 1] and on it let the equilateral triangle DAB be constructed.

连接 AB，并在 AB 上作等边三角形 DAB（公设 1，euclid-elements/book1-prop-001）。

[I. 1] Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2] with centre B and distance BC let the circle CGH be described; [Post. 3] and again, with centre D and distance DG let the circle GKL be described. [Post. 3] Then, since the point B is the centre of the circle CGH, BC is equal to BG.

把 DA、DB 延长；以 B 为圆心、BC 为距离作圆，取交点 G；再以 D 为圆心、DG 为距离作圆，取 DA 延长线上的 L。

Again, since the point D is the centre of the circle GKL, DL is equal to DG. And in these DA is equal to DB; therefore the remainder AL is equal to the remainder BG. [C.N. 3] But BC was also proved equal to BG; therefore each of the straight lines AL, BC is equal to BG.

由圆的定义，BG 等于 BC，DL 等于 DG；由等边三角形，DA 等于 DB。

And things which are equal to the same thing are also equal to one another; [C.N. 1] therefore AL is also equal to BC. Therefore at the given point A the straight line AL is placed equal to the given straight line BC.

从相等的 DL、DG 中减去相等的 DA、DB，得 AL 等于 BG；又 BG 等于 BC，所以从 A 作得 AL 等于 BC。