If a straight line is cut in extreme and mean ratio, and a straight line equal to the greater segment is added to it, then the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment.

Let the straight line *AB* be cut in extreme and mean ratio at the point *C,* let *AC* be the greater segment, and let *AD* be equal to *AC.*

I say that the straight line *DB* is cut in extreme and mean ratio at *A,* and the original straight line *AB* is the greater segment.

Describe the square *AE* on *AB,* and draw the figure.

Since *AB* is cut in extreme and mean ratio at *C,* therefore the rectangle *AB* by *BC* equals the square on *AC.*

And *CE* is the rectangle *AB* by *BC,* and *CH* is the square on *AC,* therefore *CE* equals *HC.*

But *HE* equals *CE,* and *DH* equals *HC,* therefore *DH* also equals *HE.* Therefore the whole *DK* is equal to the whole *AE.*

And *DK* is the rectangle *BD* by *DA,* for *AD* equals *DL,* and *AE* is the square on *AB,* therefore the rectangle *BD* by *DA* equals the square on *AB.*

Therefore *DB* is to *BA* as *BA* is to *AD.* And *DB* is greater than *BA,* therefore *BA* is also greater than *AD.*

Therefore *DB* has been cut in extreme and mean ratio at *A,* and *AB* is the greater segment.

Therefore, *if a straight line is cut in extreme and mean ratio, and a straight line equal to the greater segment is added to it, then the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment.*

Q.E.D.