If the square on a straight line is five times the square on a segment on it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line.

Let the square on the straight line *AB* be five times the square on the segment *AC* of it, and let *CD* be double *AC.*

I say that, when *CD* is cut in extreme and mean ratio, then the greater segment is *CB.*

Describe the squares *AF* and *CG* on *AB* and *CD* respectively, draw the figure in *AF,* and draw *BE* through.

Now, since the square on *BA* is five times the square on *AC,* therefore *AF* is five times *AH.* Therefore the gnomon *MNO* is quadruple *AH.*

And, since *DC* is double *CA,* therefore the square on *DC* is quadruple the square on *CA,* that is, *CG* is quadruple *AH.* But the gnomon *MNO* is also quadruple *AH,* therefore the gnomon *MNO* equals *CG.*

And, since *DC* is double *CA,* while *DC* equals *CK,* and *AC* equals *CH,* therefore *KB* is also double *BH.*

But the sum of *LH* and *HB* is also double *HB,* therefore *KB* equals the sum of *LH* and *HB.*

But the whole gnomon *MNO* was also proved equal to the whole *CG,* therefore the remainder *HF* equals *BG.*

And *BG* is the rectangle *CD* by *DB,* for *CD* equals *DG,* and *HF* is the square on *CB,* therefore the rectangle *CD* by *DB* equals the square on *CB.*

Therefore *DC* is to *CB* as *CB* is to *BD.* But *DC* is greater than *CB,* therefore *CB* is also greater than *BD.*

Therefore, when the straight line *CD* is cut in extreme and mean ratio, *CB* is the greater segment.

Q.E.D.

That the double *AC* is greater than *BC* is to be proved thus.

If not, let *BC* be, if possible, double *CA.*

Therefore the square on *BC* is quadruple the square on *CA.* Therefore the sum of the squares on *BC* and *CA* is five times the square on *CA.* But, by hypothesis, the square on *BA* is also five times the square on *CA*

Therefore the square on *BA* equals the sum of the squares on *BC* and *CA,* which is impossible.

Therefore *CB* is not double *AC.*

Similarly we can prove that neither is a straight line less than *CB* double *CA,* for the absurdity is much greater.

Therefore the double *AC* is greater than *CB.*

Therefore, *if the square on a straight line is five times the square on a segment on it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line.*

Q.E.D.