II.9 Proposition II.9
If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section.
Proof
Let AB be bisected at C (I.10) and cut unequally at D. At C
draw CE at right angles to AB (I.11), with CE=AC=CB. Join
EA and EB; through D draw DF parallel to CE (I.31), meeting
EB at F; through F draw FG parallel to AB (I.31), meeting
CE at G. Join AF.
Since ∠ECA is a right angle and CA=CE, the triangle
ACE is right-isoceles, and ∠CAE=∠AEC= half a right
angle (I.5; I.32). Similarly ∠CBE=∠CEB= half a
right angle. Hence ∠AEB is a right angle (Common Notion 2),
and triangle AEB is right-angled at E. By I.47:
\[
AB^2 \;=\; AE^2 + EB^2.
\]
Now AE2=AC2+CE2=2⋅AC2 (I.47 in △ACE,
plus CE=AC). Similarly inside the right triangles formed by the
perpendicular DF at D on AB, I.47 plus the fact that
DF=DE (which one shows from the parallel construction)
yields, after Common Notions 2 and 3:
\[
AD^2 + DB^2 \;=\; 2\cdot AC^2 + 2\cdot CD^2.
\]
lines 74–74 in main.tex
