| If a straight line is bisected and a straight line is added to it in a straight line, then the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half equals the square on the straight line made up of the half and the added straight line. | ||
| 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 AD by DB together with the square on CB equals the square on CD. | ||
| Describe the square CEFD on CD, and join DE. Draw BG through the point B parallel to either EC or DF, draw KM through the point H parallel to either AB or EF, and further draw AK through A parallel to either CL or DM. | I.46 | |
| Then, since AC equals CB, AL also equals CH. But CH equals HF. Therefore AL also equals HF. | I.36
I.43 | |
| Add CM to each. Therefore the whole AM equals the gnomon NOP. | II.Def.2 | |
| But AM is the rectangle AD by DB, for DM equals DB. Therefore the gnomon NOP also equals the rectangle AD by DB. | ||
| Add LG, which equals the square on BC, to each. Therefore the rectangle AD by DB together with the square on CB equals the gnomon NOP plus LG. | ||
| But the gnomon NOP and LG are the whole square CEFD, which is described on CD. | ||
| Therefore the rectangle AD by DB together with the square on CB equals the square on CD. | ||
| Therefore if a straight line is bisected and a straight line is added to it in a straight line, then the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half equals the square on the straight line made up of the half and the added straight line. | ||
| Q.E.D. | ||
