For let the straight line AB be cut in extreme and mean ratio at
the point C, and let AC be the greater segment; let the straight
line AD be produced in a straight line with CA, and let AD be made
half of AB; I say that the square on CD is five times the square
on AD.
For let the squares AE, DF be described on AB, DC, and let the figure
in DF be drawn; let FC be carried through to G.
Now, since AB has been cut in extreme and mean ratio at C, therefore
the rectangle AB, BC is equal to the square on AC. [VI. Def. 3,
VI. 17]
And CE is the rectangle AB, BC, and FH the square on AC; therefore
CE is equal to FH.
And, since BA is double of AD, while BA is equal to KA, and AD to
AH, therefore KA is also double of AH.
But, as KA is to AH, so is CK to CH; [VI. 1] therefore CK is double
of CH.
But LH, HC are also double of CH.
Therefore KC is equal to LH, HC. [p. 441]
But CE was also proved equal to HF; therefore the whole square AE
is equal to the gnomon MNO.
And, since BA is double of AD, the square on BA is quadruple of
the square on AD, that is, AE is quadruple of DH.
But AE is equal to the gnomon MNO; therefore the gnomon MNO is also
quadruple of AP; therefore the whole DF is five times AP.
And DF is the square on DC, and AP the square on DA; therefore the
square on CD is five times the square on DA.
Therefore etc.
Q. E. D.
