Proof: By Euclid

For let the area $ABCD$ be contained by the rational (straight line) $AB$ and the sixth binomial (straight line) $AD$, which has been divided into its (component) terms at $E$, such that $AE$ is the greater term.
So, I say that the square root of $AC$ is the square root of (the sum of) two medial (areas).

fig054e

For let the same construction be made as that shown previously.
So, (it is) clear that $MO$ is the square root of $AC$, and that $MN$ is incommensurable in square with $NO$.
And since $EA$ is incommensurable in length with $AB$ [Def. 10.10] , $EA$ and $AB$ are thus rational (straight lines which are) commensurable in square only.
Thus, $AK$ - that is to say, the sum of the (squares) on $MN$ and $NO$ - is medial [Prop. 10.21].
Again, since $ED$ is incommensurable in length with $AB$ [Def. 10.10] , $FE$ is thus also incommensurable (in length) with $EK$ [Prop. 10.13].
Thus, $FE$ and $EK$ are rational (straight lines which are) commensurable in square only.
Thus, $EL$ - that is to say, $MR$ - that is to say, the (rectangle contained) by $MNO$ - is medial [Prop. 10.21].
And since $AE$ is incommensurable (in length) with $EF$, $AK$ is also incommensurable with $EL$ [Prop. 6.1], [Prop. 10.11].
But, $AK$ is the sum of the (squares) on $MN$ and $NO$, and $EL$ is the (rectangle contained) by $MNO$.
Thus, the sum of the (squares) on $MNO$ is incommensurable with the (rectangle contained) by $MNO$.
And each of them is medial.
And $MN$ and $NO$ are incommensurable in square.
Thus, $MO$ is the square root of (the sum of) two medial (areas) [Prop. 10.41].
And (it is) the square root of $AC$.
(Which is) the very thing it was required to show.
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