第1卷命题 1 · 在给定有限直线上作等边三角形

Let AB be the given finite straight line. Thus it is required to construct an equilateral triangle on the straight line AB. With centre A and distance AB let the circle BCD be described; [Post. 3] again, with centre B and distance BA let the circle ACE be described; [Post. 3] and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined.

以 A 为圆心、AB 为半径作圆，再以 B 为圆心、BA 为半径作圆；两圆交于 C，连接 AC 与 BC（公设 1、3）。

[Post. 1] Now, since the point A is the centre of the circle CDB, AC is equal to AB. [Def. 15] Again, since the point B is the centre of the circle CAE, BC is equal to BA.

因为 A 是第一圆圆心，AC 等于 AB；因为 B 是第二圆圆心，BC 等于 BA。

[Def. 15] But CA was also proved equal to AB; therefore each of the straight lines CA, CB is equal to AB. And things which are equal to the same thing are also equal to one another; [C.N. 1] therefore CA is also equal to CB.

AB 与 BA 是同一条给定线段，所以 AC 与 BC 都等于 AB；由公理 1，AC 等于 BC。

Therefore the three straight lines CA, AB, BC are equal to one another. Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB.

于是 AB、AC、BC 三边相等，三角形 ABC 是等边三角形。