If a straight line is cut at random, then the sum of the rectangles contained by the whole and each of the segments equals the square on the whole.

Let the straight line *AB* be cut at random at the point *C.*

I say that the sum of the rectangle *BA* by *AC* and the rectangle *AB* by *BC* equals the square on *AB*

Describe the square *ADEB* on *AB,* and draw *CF* through *C* parallel to either *AD* or *BE.*
Then *AE* equals *AF* plus *CE.*

Now *AE* is the square on *AB*; *AF* is the rectangle *BA* by *AC,* for it is contained by *DA* and *AC,* and *AD* equals *AB*; and *CE* is the rectangle *AB* by *BC,* for *BE* equals *AB.*

Therefore the sum of the rectangle *BA* by *AC* and the rectangle *AB* by *BC* equals the square on *AB.*

Therefore *if a straight line is cut at random, then the sum of the rectangles contained by the whole and each of the segments equals the square on the whole.*

Q.E.D.

In modern algebraic notation this proposition says that
if *y* = *y*_{1} + *y*_{2}, then *xy* = *x* *y*_{1} + *x* *y*_{2}. This can also be stated in a single equation as

This proposition is used in the proof of proposition XIII.10 which shows that a certain relationship holds for the sides of a regular pentagon, regular hexagon, and regular decagon that are all inscribed in the same circle, namely, the square on the side of the pentagon equals the sum of the squares on the side of a hexagon and on the side of a decagon.