Let AB be a straight line, let it be cut in extreme and mean ratio
at C, and let AC be the greater segment; I say that the squares
on AB, BC are triple of the square on CA.

For let the square ADEB be described on AB, and let the figure
be drawn.

Since then AB has been cut in extreme and mean ratio at C, and AC
is the greater segment, therefore the rectangle AB, BC is equal
to the square on AC. [VI. Def. 3, VI. 17]

And AK is the rectangle AB, BC, and HG the square on AC; therefore
AK is equal to HG. [p. 448]

And, since AF is equal to FE, let CK be added to each; therefore
the whole AK is equal to the whole CE; therefore AK, CE are double
of AK.

But AK, CE are the gnomon LMN and the square CK; therefore the gnomon
LMN and the square CK are double of AK.

But, further, AK was also proved equal to HG; therefore the gnomon
LMN and the squares CK, HG are triple of the square HG.

And the gnomon LMN and the squares CK, HG are the whole square AE
and CK, which are the squares on AB, BC, while HG is the square
on AC.

Therefore the squares on AB, BC are triple of the square on AC.

Q. E. D.