If a straight line is cut into equal and unequal segments, then the sum of the squares on the unequal segments of the whole is double the sum of the square on the half and the square on the straight line between the points of section.

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

I say that the sum of the squares on *AD* and *DB* is double the sum of the squares on *AC* and *CD.*

Draw *CE* from *C* at right angles to *AB,* and make it equal to either *AC* or *CB.* Join *EA* and *EB.* Draw *DF* through *D* parallel to *EC* and *FG* through *F* parallel to *AB.* Join *AF.*

Then, since *AC* equals *CE,* the angle *EAC* also equals the angle *AEC.*

And, since the angle at *C* is right, the sum of the remaining angles *EAC* and *AEC* equals one right angle.

And they are equal, therefore each of the angles *CEA* and *CAE* is half of a right angle.

For the same reason each of the angles *CEB* and *EBC* is also half of a right angle, therefore the whole angle *AEB* is right.

And, since the angle *GEF* is half of a right angle, and the angle *EGF* is right, for it equals the interior and opposite angle *ECB,* the remaining angle *EFG* is half of a right angle. Therefore the angle *GEF* equals the angle *EFG,* so that the side *EG* also equals *GF.*

Again, since the angle at *B* is half of a right angle, and the angle *FDB* is right, for it is again equal to the interior and opposite angle *ECB,* the remaining angle *BFD* is half of a right angle. Therefore the angle at *B* equals the angle *DFB,* so that the side *FD* also equals the side *DB.*

Now, since *AC* equals *CE,* the square on *AC* also equals the square on *CE,* therefore the sum of the squares on *AC* and *CE* is double the square on *AC.*

But the square on *EA* equals the sum of the squares on *AC* and *CE,* for the angle *ACE* is right, therefore the square on *EA* is double the square on *AC.*

Again, since *EG* equals *GF,* the square on *EG* also equals the square on *GF.* Therefore the sum of the squares on *EG* and *GF* is double the square on *GF.*

But the square on *EF* equals the sum of the squares on *EG* and *GF,* therefore the square on *EF* is double the square on *GF.*

But *GF* equals *CD,* therefore the square on *EF* is double the square on *CD.*

But the square on *EA* is also double of the square on *AC,* therefore the sum of the squares on *AE* and *EF* is double the sum of the squares on *AC* and *CD.*

And the square on *AF* equals sum of the squares on *AE* and *EF,* for the angle *AEF* is right. Therefore the square on *AF* is double the sum of the squares on *AC* and *CD.*

But the sum of the squares on *AD* and *DF* equals the square on *AF,* for the angle at *D* is right, therefore the sum of the squares on *AD* and *DF* is double the sum the squares on *AC* and *CD.*

And *DF* equals *DB.*

Therefore the sum of the squares on *AD* and *DB* is double the sum of the squares on *AC* and *CD.*

Therefore *if a straight line is cut into equal and unequal segments, then the sum of the squares on the unequal segments of the whole is double the sum of the square on the half and the square on the straight line between the points of section.*

Q.E.D.

Draw *CE* at right angles to *AB* and equal to half of it. Finish the diagram by drawing parallel lines and connecting points. That results in four isosceles right triangles, *ACE, ECB, EGF,* and *FDB,*
as well as two other right triangles, *AEB* and *AEF.* Then,

AF^{2} | = | AE^{2} + EF^{2} |

= | (AC^{2} + CE^{2}) +
(EG^{2} + GF^{2}) | |

= | 2(AC^{2}+CD^{2}) |

AF^{2} | = | AD^{2} + DF^{2} |

= | AD^{2} + DB^{2} |

Thus, *AD*^{2} + *DB*^{2} = 2(*AC*^{2}+*CD*^{2}), as desired.

Up until this proposition, Euclid has only used cut-and-paste proofs, and such a proof can be made for this proposition as well. It would start with the same line AB bisected at C and also cut at D. Then lines at right angles and parallel to line AB would be constructed to make squares and rectangles of various sizes.
The goal is to show that AD^{2} + DB^{2} = 2(AC^{2} + CD^{2}).
The left hand side of the equation is displayed to the right as the two squares |

For instance, when *AC* and *CB* are set to *y* while *CD* is set to *z,* then the algebraic identity

results.

Alternatively, when *AC* and *CB* are set to *y* while *BC* is set to *x,*, then we get the identity

There are yet other interpretations. This proposition is used in Book X to prove a lemma for X.60, and in that lemma, *w* = *AD* and *x* = *DB* are given first, so that