If a straight line be bisected and a straight line be added to it in a straight line, the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and of the square described on the straight line made up of the half and the added straight line as on one straight line.
Proof
Let AB be bisected at C (I.10) and produced to D. At C erect
CE at right angles to AB (I.11), with CE=CA=CB. Join EA,
EB, ED. Through D draw DF parallel to CE, of length such
that DF meets the line through E parallel to AB at F (I.31).
As in II.9, ∠AEB is a right angle (right-isoceles triangles
ACE and BCE). Triangle ADE is also right-angled, with the
right angle at E in the configuration where D lies on the
extension of AB. Apply I.47 twice (to △AED and to the
triangle formed by extending the constructions to F):
\[
AD^2 + DB^2 \;=\; 2\cdot AC^2 + 2\cdot CD^2,
\]
where now CD=CB+BD is the half plus the added segment. The
derivation parallels II.9 exactly with extension in place of internal
cut, and the symmetry is what Heath emphasises.