that is to GP; [II. 4, Corollary.
therefore CG is equal to GP. [Axiom 1.
And because CG is equal to GP, and PR to RO, the rectangle AG is equal to the rectangle MP, and the rectangle PL to the rectangle RF, [I. 36.
But MP is equal to PL, because they are the complements of the parallelogram ML; [I. 43.
therefore also AG is equal to RF. [Axiom 1.
Therefore the four rectangles AG, MP, PL, RF are equal to one another, and so the four are quadruple of one of them AG.
And it was shewn that the four CK, BN, GR and RN are quadruple of CK; therefore the eight rectangles which make up the gnomon AOH are quadruple of AK.
And because AK is the rectangle contained by AB, BC; for BK is equal to BC;
therefore four times the rectangle AB, BC is quadruple of AK.
But the gnomon AOH was shewn to be quadruple of AK,
Therefore four times the rectangle AB, BC is equal to the gnomon AOH. [Axiom 1.
To each of these add XH, which is equal to the square on AC. [II. 4, Corollary, and I. 34.
Therefore four times the rectangle AB, BC, together with the square on AC, is equal to the gnomon AOH and the square XH.
But the gnomon AOH and the square XH make up the figure AEFD, which is the square on AD.
Therefore four times the rectangle AB, BC, together with the square on AC, is equal to the square on AD, that is to the square on the line made of AB and BC together.
Wherefore, if a straight line &c. q.e.d.
