Proof: By Euclid

Let $AB$ and $C$ be the two given unequal straight lines, and let $AB$ be the greater of them.
So it is required to find by (the square on) which (straight line) the square on $AB$ (is) greater than (the square on) $C$.

fig013ae

Let the semicircle $ADB$ have been described on $AB$.
And let $AD$, equal to $C$, have been inserted into it [Prop. 4.1].
And let $DB$ have been joined.
So (it is) clear that the angle $ADB$ is a right angle [Prop. 3.31], and that the square on $AB$ (is) greater than (the square on) $AD$ - that is to say, (the square on) $C$ - by (the square on) $DB$ [Prop. 1.47].
And, similarly, the square root of (the sum of the squares on) two given straight lines is also found likeso.
Let $AD$ and $DB$ be the two given straight lines.
And let it be necessary to find the square root of (the sum of the squares on) them.
For let them have been laid down such as to encompass a right angle - (namely), that (angle encompassed) by $AD$ and $DB$.
And let $AB$ have been joined.
(It is) again clear that $AB$ is the square root of (the sum of the squares on) $AD$ and $DB$ [Prop. 1.47].
(Which is) the very thing it was required to show.
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