If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.
若一条直线被平分，并在一直线上加上一段，则整线连同所加段与所加段所成矩形，加半线上的正方形，等于半线加所加段上的正方形。
AB 在 C 处平分，再延长 BD。CEFD 是建于 CD 上的正方形，DE 为对角线。过 B 作 BG⊥CD 交 EF 于 G；过 H（BG 与 DE 的交点）作 KM∥CD；过 A 作 AK∥CE。LG 为 BC 上的正方形，gnomon NOP 围绕 LG。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
For let a straight line AB be bisected at the point C, and let a straight line BD be added to it in a straight line; I say that the rectangle contained by AD, DB together with the square on CB is equal to the square on CD. For let the square CEFD be described on CD, [I. 46] and let DE be joined; through the point B let BG be drawn parallel to either EC or DF, through the point H let KM be drawn parallel to either AB or EF, and further through A let AK be drawn parallel to either CL or DM. [I. 31] Then, since AC is equal to CB, AL is also equal to CH.
把给定线平分，并在同一直线上加上一段。
[I. 36] But CH is equal to HF. [I. 43] Therefore AL is also equal to HF.
以“半线加延长段”为边作正方形。
Let CM be added to each; therefore the whole AM is equal to the gnomon NOP. But AM is the rectangle AD, DB, for DM is equal to DB; therefore the gnomon NOP is also equal to the rectangle AD, DB.
按原半点和原端点引平行线，分解出整线连同延长段与延长段所成矩形，以及半线平方。
Let LG, which is equal to the square on BC, be added to each; therefore the rectangle contained by AD, DB together with the square on CB is equal to the gnomon NOP and LG. But the gnomon NOP and LG are the whole square CEFD, which is described on CD; therefore the rectangle contained by AD, DB together with the square on CB is equal to the square on CD.
这些部分合成所作正方形，得到命题关系。