To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.
作一个平行四边形，使它等于给定直线形，并有一个角等于给定直线角。
给定直线形 ABCD 化为给定角 E 的平行四边形：先三角剖分，再依次贴上 FGHK、HKLM 等等积小平行四边形，得到总平行四边形 FGLM。
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 the given rectilineal figure and E the given rectilineal angle; thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD. Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I. 42] let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I. 44] Then, since the angle E is equal to each of the angles HKF, GHM, the angle HKF is also equal to the angle GHM. [C.N. 1] Let the angle KHG be added to each; therefore the angles FKH, KHG are equal to the angles KHG, GHM.
把给定直线形分割成若干三角形。
But the angles FKH, KHG are equal to two right angles; [I. 29] therefore the angles KHG, GHM are also equal to two right angles. Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles; therefore KH is in a straight line with HM. [I. 14] And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another.
对第一个三角形用 euclid-elements/book1-prop-044 作等面积且有给定角的平行四边形。
[I. 29] Let the angle HGL be added to each; therefore the angles MHG, HGL are equal to the angles HGF, HGL. [C.N. 2] But the angles MHG, HGL are equal to two right angles; [I. 29] therefore the angles HGF, HGL are also equal to two right angles. [C.N. 1] Therefore FG is in a straight line with GL. [I. 14] And, since FK is equal and parallel to HG, [I. 34] and HG to ML also, KF is also equal and parallel to ML; [C.N. 1; I. 30] and the straight lines KM, FL join them (at their extremities); therefore KM, FL are also equal and parallel.
依次把其余三角形转成同角平行四边形并贴合。
[I. 33] Therefore KFLM is a parallelogram. And, since the triangle ABD is equal to the parallelogram FH, and DBC to GM, the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E.
合成后的平行四边形等于整个给定直线形。