To bisect a given rectilineal angle.
给定一个直线角，作一条线把它平分。
角 BAC 顶点 A，两边到 B、C；D 在 AB 上、E 在 AC 上使 AD=AE；在 DE 上作等边三角形 DEF，AF 平分该角。
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 the angle BAC be the given rectilineal angle. Thus it is required to bisect it.
在角的两边上取相等的 AD、AE，连接 DE。
Let a point D be taken at random on AB; let AE be cut off from AC equal to AD; [I. 3] let DE be joined, and on DE let the equilateral triangle DEF be constructed; let AF be joined. I say that the angle BAC has been bisected by the straight line AF.
在 DE 上作等边三角形，并连接顶点到原角顶点。
For, since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively. And the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF.
两个小三角形有对应三边相等，由 euclid-elements/book1-prop-008 对应角相等。
[I. 8] Therefore the given rectilineal angle BAC has been bisected by the straight line AF.
因此从原角顶点引出的这条线平分给定角。