On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle.
在给定直线的一点上，作一个直线角，使它等于给定直线角。
给定角 DCE 与直线 AB 上的点 A。在 CD、CE 上取 D、E 连成三角形 CDE；以 [[I.22]] 在 AB 上 A 处作三角形 AFG 与 CDE 全等，从而角 FAG 等于角 DCE。
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 straight line, A the point on it, and the angle DCE the given rectilineal angle; thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE.
在给定角上取两条边段，并连接成一个三角形。
On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined, and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed in such a way that CD is equal to AF, CE to AG, and further DE to FG.
在给定点所在直线上，用 euclid-elements/book1-prop-022 作一个三角形，使对应三边相等。
Then, since the two sides DC, CE are equal to the two sides FA, AG respectively, and the base DE is equal to the base FG, the angle DCE is equal to the angle FAG.
由 euclid-elements/book1-prop-008，新三角形的对应角等于原角。
[I. 8] Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE.
所以在给定点作得一个等于给定角的角。