Proof: By Euclid
(related to Proposition: 2.06: Square of Sum with One Halved Summand)
 For let the square $CEFD$ have been described on $CD$ [Prop. 1.46], and let $DE$ have been joined, and let $BG$ have been drawn through point $B$, parallel to either of $EC$ or $DF$ [Prop. 1.31], and let $KM$ have been drawn through point $H$, parallel to either of $AB$ or $EF$ [Prop. 1.31], and finally let $AK$ have been drawn through $A$, parallel to either of $CL$ or $DM$ [Prop. 1.31].
 Therefore, since $AC$ is equal to $CB$, (rectangle) $AL$ is also equal to (rectangle) $CH$ [Prop. 1.36].
 But, (rectangle) $CH$ is equal to (rectangle) $HF$ [Prop. 1.43].
 Thus, (rectangle) $AL$ is also equal to (rectangle) $HF$.
 Let (rectangle) $CM$ have been added to both.
 Thus, the whole (rectangle) $AM$ is equal to the gnomon $NOP$.
 But, $AM$ is the (rectangle contained) by $AD$ and $DB$.
 For $DM$ is equal to $DB$.
 Thus, gnomon $NOP$ is also equal to the [rectangle contained] by $AD$ and $DB$.
 Let $LG$, which is equal to the square on $BC$, have been added to both.
 Thus, the rectangle contained by $AD$ and $DB$, plus the square on $CB$, is equal to the gnomon $NOP$ and the (square) $LG$.
 But the gnomon $NOP$ and the (square) $LG$ is (equivalent to) the whole square $CEFD$, which is on $CD$.
 Thus, the rectangle contained by $AD$ and $DB$, plus the square on $CB$, is equal to the square on $CD$.
 Thus, if a straight line is cut in half, and any straight line added to it straighton, then the rectangle contained by the whole (straight line) with the (straight line) having being added, and the (straight line) having being added, plus the square on half (of the original straight line), is equal to the square on the sum of half (of the original straight line) and the (straight line) having been added.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"