II.8 Proposition II.8
If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line.
Proof
Let AB be cut at C, and produce AB to D so that BD=BC.
Then AD=AB+BC and AD is cut at B into the segments AB
and BD=BC. Apply II.4 to the line AD cut at B: the square
on AD equals the squares on AB and BD together with twice the
rectangle on AB, BD. Since BD=BC, this becomes:
\[
AD^2 \;=\; AB^2 + BC^2 + 2\cdot(AB\cdot BC).
\]
Apply II.4 again to AB cut at C, namely AB2=AC2+CB2+2⋅(AC⋅CB), and substitute. Combining (Common Notion 2)
and re-arranging (Common Notion 3) to isolate 4⋅(AB⋅BC)
on the right side yields:
\[
AD^2 \;=\; 4\cdot(AB\cdot BC) + AC^2,
\]
which is (a+b)2=4ab+(a−b)2 in the form Euclid states it.
lines 74–74 in main.tex
