# Proof: By Euclid

(related to Proposition: 2.04: Square of Sum)

• For let the square $ADEB$ have been described on $AB$ [Prop. 1.46], and let $BD$ have been joined, and let $CF$ have been drawn through $C$, parallel to either of $AD$ or $EB$ [Prop. 1.31], and let $HK$ have been drawn through $G$, parallel to either of $AB$ or $DE$ [Prop. 1.31].
• And since $CF$ is parallel to $AD$, and $BD$ has fallen across them, the external angle $CGB$ is equal to the internal and opposite (angle) $ADB$ [Prop. 1.29].
• But, $ADB$ is equal to $ABD$, since the side $BA$ is also equal to $AD$ [Prop. 1.5].
• Thus, angle $CGB$ is also equal to $GBC$.
• So the side $BC$ is equal to the side $CG$ [Prop. 1.6].
• But, $CB$ is equal to $GK$, and $CG$ to $KB$ [Prop. 1.34].
• Thus, $GK$ is also equal to $KB$.
• Thus, $CGKB$ is equilateral.
• So I say that (it is) also right-angled.
• For since $CG$ is parallel to $BK$ [and the straight line $CB$ has fallen across them], the angles $KBC$ and $GCB$ are thus equal to two right angles [Prop. 1.29].
• But $KBC$ (is) a right angle.
• Thus, $BCG$ (is) also a right angle.
• So the opposite (angles) $CGK$ and $GKB$ are also right angles [Prop. 1.34].
• Thus, $CGKB$ is right-angled.
• And it was also shown (to be) equilateral.
• Thus, it is a square.
• And it is on $CB$.
• So, for the same (reasons), $HF$ is also a square.
• And it is on $HG$, that is to say [on] $AC$ [Prop. 1.34].
• Thus, the squares $HF$ and $KC$ are on $AC$ and $CB$ (respectively).
• And the (rectangle) $AG$ is equal to the (rectangle) $GE$ [Prop. 1.43].
• And $AG$ is the (rectangle contained) by $AC$ and $CB$.
• For $GC$ (is) equal to $CB$.
• Thus, $GE$ is also equal to the (rectangle contained) by $AC$ and $CB$.
• Thus, the (rectangles) $AG$ and $GE$ are equal to twice the (rectangle contained) by $AC$ and $CB$.
• And $HF$ and $CK$ are the squares on $AC$ and $CB$ (respectively).
• Thus, the four (figures) $HF$, $CK$, $AG$, and $GE$ are equal to the (sum of the) squares on $AC$ and $BC$, and twice the rectangle contained by $AC$ and $CB$.
• But, the (figures) $HF$, $CK$, $AG$, and $GE$ are (equivalent to) the whole of $ADEB$, which is the square on $AB$.
• Thus, the square on $AB$ is equal to the (sum of the) squares on $AC$ and $CB$, and twice the rectangle contained by $AC$ and $CB$.
• Thus, if a straight line is cut at random then the square on the whole (straight line) is equal to the (sum of the) squares on the pieces (of the straight line), and twice the rectangle contained by the pieces.
• (Which is) the very thing it was required to show.

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