Proof: By Euclid
(related to Proposition: 2.05: Rectangle is Difference of Two Squares)
 For let the square $CEFB$ have been described on $CB$ [Prop. 1.46], and let $BE$ have been joined, and let $DG$ have been drawn through $D$, parallel to either of $CE$ or $BF$ [Prop. 1.31], and again let $KM$ have been drawn through $H$, parallel to either of $AB$ or $EF$ [Prop. 1.31], and again let $AK$ have been drawn through $A$, parallel to either of $CL$ or $BM$ [Prop. 1.31].
 And since the complement $CH$ is equal to the complement $HF$ [Prop. 1.43], let the (square) $DM$ have been added to both.
 Thus, the whole (rectangle) $CM$ is equal to the whole (rectangle) $DF$.
 But, (rectangle) $CM$ is equal to (rectangle) $AL$, since $AC$ is also equal to $CB$ [Prop. 1.36].
 Thus, (rectangle) $AL$ is also equal to (rectangle) $DF$.
 Let (rectangle) $CH$ have been added to both.
 Thus, the whole (rectangle) $AH$ is equal to the gnomon $NOP$.
 But, $AH$ is the (rectangle contained) by $AD$ and $DB$.
 For $DH$ (is) equal to $DB$.
 Thus, the gnomon $NOP$ is also equal to the (rectangle contained) by $AD$ and $DB$.
 Let $LG$, which is equal to the (square) on $CD$, have been added to both.
 Thus, the gnomon $NOP$ and the (square) $LG$ are equal to the rectangle contained by $AD$ and $DB$, and the square on $CD$.
 But, the gnomon $NOP$ and the (square) $LG$ is (equivalent to) the whole square $CEFB$, which is on $CB$.
 Thus, the rectangle contained by $AD$ and $DB$, plus the square on $CD$, is equal to the square on $CB$.
 Thus, if a straight line is cut into equal and unequal (pieces) then the rectangle contained by the unequal pieces of the whole (straight line), plus the square on the (difference) between the (equal and unequal) pieces, is equal to the square on half (of the straight line).
 (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"