If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half.
若一条直线既被等分又被不等分，则不等两段所成矩形加两分点间线段上的正方形，等于半线上的正方形。
AB 被 C 等分、被 D 不等分。CEFB 是建于 CB 上的正方形，BE 为对角线。过 D 作 DG⊥AB 交 EF 于 G，交 BE 于 H；过 H 作 KM∥AB；过 A 作 AK∥CE。LG 为 CD 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 cut into equal segments at C and into unequal segments at D; I say that the rectangle contained by AD, DB together with the square on CD is equal to the square on CB. For let the square CEFB be described on CB, [I. 46] and let BE be joined; through D let DG be drawn parallel to either CE or BF, through H again let KM be drawn parallel to either AB or EF, and again through A let AK be drawn parallel to either CL or BM.
一条线既被等分又被不等分，比较半线正方形与不等两段矩形。
[I. 31] Then, since the complement CH is equal to the complement HF, [I. 43] let DM be added to each; therefore the whole CM is equal to the whole DF. But CM is equal to AL, since AC is also equal to CB; [I. 36] therefore AL is also equal to DF.
以半线为边作正方形，再用不等分点形成的偏移线切割。
Let CH be added to each; therefore the whole AH is equal to the gnomon NOP. But AH is the rectangle AD, DB, for DH is equal to DB, therefore the gnomon NOP is also equal to the rectangle AD, DB.
切出的缺口与补上的小正方形正好对应。
Let LG, which is equal to the square on CD, be added to each; therefore the gnomon NOP and LG are equal to the rectangle contained by AD, DB and the square on CD. But the gnomon NOP and LG are the whole square CEFB, which is described on CB; therefore the rectangle contained by AD, DB together with the square on CD is equal to the square on CB.
于是不等两段矩形加两分点间线段平方，等于半线平方。