If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment together with the square on the remaining segment.
Proof
Let AB be cut at C. Describe on AB the square ADEB (I.46),
and through C draw CF parallel to AD or BE (I.31), meeting
DE at F. On CB as side construct the square CBKG inside
ADEB (a copy of II.4's construction), with HK parallel to AB.
By II.4 the square ADEB on AB equals the square on AC plus the
square on CB plus twice the rectangle AC⋅CB. Add the
square on CB to both sides of this identity (Common Notion 2):
\[
AB^2 + CB^2 \;=\; AC^2 + 2\cdot CB^2 + 2\cdot(AC\cdot CB).
\]
But 2⋅CB2+2⋅(AC⋅CB)=2⋅CB(CB+AC)=2⋅CB⋅AB (Definition II.1; II.1). Hence
AB2+CB2=2⋅(AB⋅CB)+AC2, as required.