If a straight line is cut at random, then the rectangle contained by the whole and one of the segments equals the sum of the rectangle contained by the segments and the square on the aforesaid segment.

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

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

Describe the square *CDEB* on *CB.* Draw *ED* through to *F,* and draw *AF* through *A* parallel to either *CD* or *BE.*

Then *AE* equals *AD* plus *CE.*

Now *AE* is the rectangle *AB* by *BC,* for it is contained by *AB* and *BE,* and *BE* equals *BC*; *AD* is the rectangle *AC* by *CB,* for *DC* equals *CB*; and *DB* is the square on *CB.*

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

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

Q.E.D.