Proof: By Euclid

Let $FE$ and $EG$ be two straight lines.
I say that as $FE$ is to $EG$, so the (square) on $FE$ (is) to the (rectangle contained) by $FE$ and $EG$.

fig021ae

For let the square $DF$ have been described on $FE$.
And let $GD$ have been completed.
Therefore, since as $FE$ is to $EG$, so $FD$ (is) to $DG$ [Prop. 6.1], and $FD$ is the (square) on $FE$, and $DG$ the (rectangle contained) by $DE$ and $EG$ - that is to say, the (rectangle contained) by $FE$ and $EG$ - thus as $FE$ is to $EG$, so the (square) on $FE$ (is) to the (rectangle contained) by $FE$ and $EG$.
And also, similarly, as the (rectangle contained) by $GE$ and $EF$ is to the (square on) $EF$ - that is to say, as $GD$ (is) to $FD$ - so $GE$ (is) to $EF$.
(Which is) the very thing it was required to show.
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