# Proof: By Euclid

• Let $AB$ be a straight line, and let it have been cut unequally at $C$, and let $AC$ be greater (than $CB$).
• I say that (the sum of) the (squares) on $AC$ and $CB$ is greater than twice the (rectangle contained) by $AC$ and $CB$. • For let $AB$ have been cut in half at $D$.
• Therefore, since a straight line has been cut into equal (parts) at $D$, and into unequal (parts) at $C$, the (rectangle contained) by $AC$ and $CB$, plus the (square) on $CD$, is thus equal to the (square) on $AD$ [Prop. 2.5].
• Hence, the (rectangle contained) by $AC$ and $CB$ is less than the (square) on $AD$.
• Thus, twice the (rectangle contained) by $AC$ and $CB$ is less than double the (square) on $AD$.
• But, (the sum of) the (squares) on $AC$ and $CB$ [is] double (the sum of) the (squares) on $AD$ and $DC$ [Prop. 2.9].
• Thus, (the sum of) the (squares) on $AC$ and $CB$ is greater than twice the (rectangle contained) by $AC$ and $CB$.
• (Which is) the very thing it was required to show.

Github: non-Github:
@Fitzpatrick