Book 2
5
Then, since the complement CH is equal to the complement HF, [I. 43] let DM be added to each; therefore the whole CM is equal to the whole DF.
5
But CM is equal to AL, since AC is also equal to CB; [I. 36] therefore AL is also equal to DF. Let CH be added to each; therefore the whole AH is equal to the gnomon NOP.
5
But AH is the rectangle AD, DB, for DH is equal to DB, therefore the gnomon NOP is also equal to the rectangle AD, DB.
5
Let LG, which is equal to the square on CD, be added to each; therefore the gnomon NOP and LG are equal to the rectangle contained by AD, DB and the square on CD.
5
But the gnomon NOP and LG are the whole square CEFB, which is described on CB; therefore the rectangle contained by AD, DB together with the square on CD is equal to the square on CB.
5
Therefore etc. Q. E. D. between the points of section, literally between the sections, the word being the same (τομή) as that used of a conic section. It will be observed that the gnomon is indicated in the figure by three separate letters and a dotted curve. This is no doubt a clearer way of showing what exactly the gnomon is than the method usual in our text-books. In this particular case the figure of the MSS. has two M's in it, the gnomon being MNΞ. I have corrected the lettering to avoid confusion.
Proposition 6.
