If a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section equals the square on the half.

Let a straight line *AB* be cut into equal segments at *C* and into unequal segments at *D.*

I say that the rectangle *AD* by *DB* together with the square on *CD* equals the square on *CB.*

Describe the square *CEFB* on *CB,* and join *BE.* Draw *DG* through *D* parallel to either *CE* or *BF,* again draw *KM* through *H* parallel to either *AB* or *EF,* and again draw *AK* through *A* parallel to either *CL* or *BM.*

Then, since the complement *CH* equals the complement *HF,* add *DM* to each. Therefore the whole *CM* equals the whole *DF.*

But *CM* equals *AL,* since *AC* is also equal to *CB.* Therefore *AL* also equals *DF.* Add *CH* to each. Therefore the whole *AH* equals the gnomon *NOP.*

But *AH* is the rectangle *AD* by *DB,* for *DH* equals *DB,* therefore the gnomon *NOP* also equals the rectangle *AD* by *DB.*

Add *LG,* which equals the square on *CD,* to each. Therefore the sum of the gnomon *NOP* and *LG* equals the sum of the rectangle *AD* by *DB* and the square on *CD.*

But the gnomon *NOP* together with *LG* is the whole square *CEFB,* which is described on *CB.*

Therefore the rectangle *AD* by *DB* together with the square on *CD* equals the square on *CB.*

Therefore *if a straight line is cut into equal and unequal segments, then the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section equals the square on the half.*

Q.E.D.

Using symbolic algebra, we can easily verify the identity

but Euclid was restricted to geometric arguments. The argument isn’t difficult. The original rectangle *AH* is the sum of the rectangles *AL* and *CH.* By proposition I.43, the rectangles *CH* and *HF* are equal. And, of course, the rectangles *AL* and *CM* are equal. Therefore, *AH* = *AL* + *CH* = *CM* + *HF* = *CB*^{2} – *LH*^{2}, as required.

In terms of the single variable *x,* this is equivalent to solving the quadratic equation, *x*(*b* – *x*) = *c*^{2}. This equation can be written in a standard form as

If *b* is represented as the line *AB* in the diagram, and with *x* = *AD* and *y* = *BD,* the first condition *x* + *y* = *b* is satisfied. This proposition says that the product *xy* equals the square on *BC* (which is *b*/2) minus the square on *CD.* Thus, the remaining condition reduces to finding *CD* so that (*b*/2)^{2} – *CD*^{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 cut a given straight line so that the rectangle contained by the unequal segments equals a given square. Thus the given square must not be greater than the square described on the half of the given straight line.
Let Then, as described above, |

This proposition is not found in the *Elements,* but a generalization is. After II.14 the given square could be replaced by any given rectilinear figure, since II.14 constructs a square equal to a given rectilinear figure. But the full generalization is not given until proposition VI.28. Not only has the given square become a general rectilinear figure, but all the rectangles and squares have been replaced by parallelograms. That requires generalizing II.4 to parallelograms. that’s done in VI.25 which constructs a parallelgram similar to a given parallelogram and equal to a given rectilinear figure. It also requires a few technical propositions to carry out the proof.