II.11 Proposition II.11
To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.
Proof
Let AB be the given straight line. Describe on AB the square
ABDC (I.46). Bisect AC at E (I.10) and join EB. Produce
CA to F in the direction of A (Postulate 2), and lay off
AF on CA produced so that EF=EB (I.3, taking EB as
the standard length). On AF describe the square FGHA (I.46);
produce GH to meet CD at K.
Then by II.6 applied to CF bisected at A with extension AF,
the rectangle on CF, FA together with the square on EA equals
the square on EF. But EF=EB, so this rectangle plus square
on EA equals the square on EB, which by I.47 (in
△ABE, right-angled at A) equals the square on EA
plus the square on AB. Subtracting the square on EA from both
sides (Common Notion 3):
\[
CF \cdot FA \;=\; AB^2.
\]
The rectangle CK on CF, FA (= CF⋅FG since FG=FA)
equals the square on AB. Subtracting the common rectangle on
FA, AH from both, the square FGHA on FA equals the rectangle
HK on HD and DK=AB−AH. Setting AH=AF on AB (point
H on AB with AH=AF) gives the desired section: AB⋅HB=AH2.
lines 74–74 in main.tex
