Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. [I. 20] Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, and B, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C. Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3] With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined; I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C. For, since the point F is the centre of the circle DKL, FD is equal to FK. But FD is equal to A; therefore KF is also equal to A. Again, since the point G is the centre of the circle LKH, GH is equal to GK. But GH is equal to C; therefore KG is also equal to C. And FG is also equal to B; therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C. Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constructed.
给定三条线段，只要任意两条合起来大于第三条，就能作出三角形，使三边分别等于这三条线段。
给定三线段 A、B、C 单独列出；底线 DE 上依次截取 DF=A、FG=B、GH=C；圆 F 半径 FD 与圆 G 半径 GH 交于 K（也称 L），即得三角形 KFG。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
先作一条线段，依次截取三条给定线段的长度。
以两个端点为圆心、以其中两条给定线段为半径作圆。
任意两边之和大于第三边保证两圆能在同侧相交。
连接交点与两端点，即得三边分别等于给定三线段的三角形。