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.*

Then, since *AC* equals *CB, AL* also equals *CH.* But *CH* equals *HF.* Therefore *AL* also equals *HF.*

Add *CM* to each. Therefore the whole *AM* equals the gnomon *NOP.*

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.

Let *b* denote the line *AB,* *x* denote *AD,* and *y* denote *BD* as in II.5. Then *x* – *y* = *b* (as opposed to *x* + *y* = *b* as in II.5).
According to this proposition the rectangle *AD* by *DB,* which is the product *xy,* is the difference of two squares, the large one being the square on the line *CD,* that is the square of *x* – *b*/2, and the small one being the square on the line *CB,* that is, the square of *b*/2. Algebraically,

This equation is easily verified with modern algebra, but it’s also easily verified in geometry, as done here in the proof.

The geometric proof is primarily an exercise in cutting and pasting. The rectangle *AB* by *DB* is the rectangle *AM,* which is the sum of the rectangles *AL* and *CM.* But the rectangles *AL, CH,* and *HF* are all equal. Therefore, the rectangle *AB* by *DB* equals the gnomon formed by the rectangles *CM* and *HF.* That gnomon is the square *CF* minus the square *LG,* but the latter equals the square on *BC.* Thus, the rectangle *AB* by *DB* equals the square on *DB* minus the square on *CB.*

a known value

In terms of *x* alone, this is equivalent to solving the quadratic equation *x*(*x* – *b*) = *c*^{2}.
Since this proposition says that *x*(*x* – *b*) = (*x* – *b*/2)^{2} – (*b*/2)^{2},
the problem reduces to solving the equation

that is, finding *CD* so that *CD*^{2} = (*b*/2)^{2} + *c*^{2}. By I.47, if a right triangle is constructed with one side equal to *b*/2 and another equal to *c,* then the hypotenuse will equal the required value for *CD.* Algebraically, the solutions *AD* for *x* and *BD* for *y* have the values

This analysis yields a construction to solve the quadratic problem stated above.

To apply a rectangle equal to a given square to a given straight line but exceeding it by a square.
Let Then, as described above, |

This construction is not found in the *Elements,* but a generalization of it to parallelograms is proposition VI.29.