# Proof: By Euclid

• For let $BD$ have been produced in a straight line [with the straight line $AB$], and let $BD$ be made equal to $CB$ [Prop. 1.3], and let the square $AEFD$ have been described on $AD$ [Prop. 1.46], and let the (rest of the) figure have been drawn double.
• Therefore, since $CB$ is equal to $BD$, but $CB$ is equal to $GK$ [Prop. 1.34], and $BD$ to $KN$ [Prop. 1.34], $GK$ is thus also equal to $KN$.
• So, for the same (reasons), $QR$ is equal to $RP$.
• And since $BC$ is equal to $BD$, and $GK$ to $KN$, (square) $CK$ is thus also equal to (square) $KD$, and (square) $GR$ to (square) $RN$ [Prop. 1.36].
• But, (square) $CK$ is equal to (square) $RN$.
• For (they are) complements in the parallelogram $CP$ [Prop. 1.43].
• Thus, (square) $KD$ is also equal to (square) $GR$.
• Thus, the four (squares) $DK$, $CK$, $GR$, and $RN$ are equal to one another.
• Thus, the four (taken together) are quadruple (square) $CK$.
• Again, since $CB$ is equal to $BD$, but $BD$ (is) equal to $BK$ - that is to say, $CG$ - and $CB$ is equal to $GK$ - that is to say, $GQ$ - $CG$ is thus also equal to $GQ$.
• And since $CG$ is equal to $GQ$, and $QR$ to $RP$, (rectangle) $AG$ is also equal to (rectangle) $MQ$, and (rectangle) $QL$ to (rectangle) $RF$ [Prop. 1.36].
• But, (rectangle) $MQ$ is equal to (rectangle) $QL$.
• For (they are) complements in the parallelogram $ML$ [Prop. 1.43].
• Thus, (rectangle) $AG$ is also equal to (rectangle) $RF$.
• Thus, the four (rectangles) $AG$, $MQ$, $QL$, and $RF$ are equal to one another.
• Thus, the four (taken together) are quadruple (rectangle) $AG$.
• And it was also shown that the four (squares) $CK$, $KD$, $GR$, and $RN$ (taken together are) quadruple (square) $CK$.
• Thus, the eight (figures taken together), which comprise the gnomon $STU$, are quadruple (rectangle) $AK$.
• And since $AK$ is the (rectangle contained) by $AB$ and $BD$, for $BK$ (is) equal to $BD$, four times the (rectangle contained) by $AB$ and $BD$ is quadruple (rectangle) $AK$.
• But the gnomon $STU$ was also shown (to be equal to) quadruple (rectangle) $AK$.
• Thus, four times the (rectangle contained) by $AB$ and $BD$ is equal to the gnomon $STU$.
• Let $OH$, which is equal to the square on $AC$, have been added to both.
• Thus, four times the rectangle contained by $AB$ and $BD$, plus the square on $AC$, is equal to the gnomon $STU$, and the (square) $OH$.
• But, the gnomon $STU$ and the (square) $OH$ is (equivalent to) the whole square $AEFD$, which is on $AD$.
• Thus, four times the (rectangle contained) by $AB$ and $BD$, plus the (square) on $AC$, is equal to the square on $AD$.
• And $BD$ (is) equal to $BC$.
• Thus, four times the rectangle contained by $AB$ and $BC$, plus the square on $AC$, is equal to the (square) on $AD$, that is to say the square described on $AB$ and $BC$, as on one (complete straight line).
• Thus, if a straight line is cut at random then four times the rectangle contained by the whole (straight line), and one of the pieces (of the straight line), plus the square on the remaining piece, is equal to the square described on the whole and the former piece, as on one (complete straight line).
• (Which is) the very thing it was required to show.

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