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Proof: By Euclid

(related to Proposition: 2.09: Sum of Squares of Sum and Difference)

• For let $CE$ have been drawn from (point) $C$, at right angles to $AB$ [Prop. 1.11], and let it be made equal to each of $AC$ and $CB$ [Prop. 1.3], and let $EA$ and $EB$ have been joined.
• And let $DF$ have been drawn through (point) $D$, parallel to $EC$ [Prop. 1.31], and (let) $FG$ (have been drawn) through (point) $F$, (parallel) to $AB$ [Prop. 1.31].
• And let $AF$ have been joined.
• And since $AC$ is equal to $CE$, the angle $EAC$ is also equal to the (angle) $AEC$ [Prop. 1.5].
• And since the (angle) at $C$ is a right angle, the (sum of the) remaining angles (of triangle $AEC$), $EAC$ and $AEC$, is thus equal to one right angle [Prop. 1.32].
• And they are equal.
• Thus, (angles) $CEA$ and $CAE$ are each half a right angle.
• So, for the same (reasons), (angles) $CEB$ and $EBC$ are also each half a right angle.
• Thus, the whole (angle) $AEB$ is a right angle.
• And since $GEF$ is half a right angle, and $EGF$ (is) a right angle - for it is equal to the internal and opposite (angle) $ECB$ [Prop. 1.29] - the remaining (angle) $EFG$ is thus half a right angle [Prop. 1.32].
• Thus, angle $GEF$ [is] equal to $EFG$.
• So the side $EG$ is also equal to the (side) $GF$ [Prop. 1.6].
• Again, since the angle at $B$ is half a right angle, and (angle) $FDB$ (is) a right angle - for again it is equal to the internal and opposite (angle) $ECB$ [Prop. 1.29] - the remaining (angle) $BFD$ is half a right angle [Prop. 1.32].
• Thus, the angle at $B$ (is) equal to $DFB$.
• So the side $FD$ is also equal to the side $DB$ [Prop. 1.6].
• And since $AC$ is equal to $CE$, the (square) on $AC$ (is) also equal to the (square) on $CE$.
• Thus, the (sum of the) squares on $AC$ and $CE$ is double the (square) on $AC$.
• And the square on $EA$ is equal to the (sum of the) squares on $AC$ and $CE$.
• For angle $ACE$ (is) a right angle [Prop. 1.47].
• Thus, the (square) on $EA$ is double the (square) on $AC$.
• Again, since $EG$ is equal to $GF$, the (square) on $EG$ (is) also equal to the (square) on $GF$.
• Thus, the (sum of the squares) on $EG$ and $GF$ is double the square on $GF$.
• And the square on $EF$ is equal to the (sum of the) squares on $EG$ and $GF$ [Prop. 1.47].
• Thus, the square on $EF$ is double the (square) on $GF$.
• And $GF$ (is) equal to $CD$ [Prop. 1.34].
• Thus, the (square) on $EF$ is double the (square) on $CD$.
• And the (square) on $EA$ is also double the (square) on $AC$.
• Thus, the (sum of the) squares on $AE$ and $EF$ is double the (sum of the) squares on $AC$ and $CD$.
• And the square on $AF$ is equal to the (sum of the squares) on $AE$ and $EF$.
• For the angle $AEF$ is a right angle [Prop. 1.47].
• Thus, the square on $AF$ is double the (sum of the squares) on $AC$ and $CD$.
• And the (sum of the squares) on $AD$ and $DF$ (is) equal to the (square) on $AF$.
• For the angle at $D$ is a right angle [Prop. 1.47].
• Thus, the (sum of the squares) on $AD$ and $DF$ is double the (sum of the) squares on $AC$ and $CD$.
• And $DF$ (is) equal to $DB$.
• Thus, the (sum of the) squares on $AD$ and $DB$ is double the (sum of the) squares on $AC$ and $CD$.
• Thus, if a straight line is cut into equal and unequal (pieces) then the (sum of the) squares on the unequal pieces of the whole (straight line) is double the (sum of the) square on half (the straight line) and (the square) on the (difference) between the (equal and unequal) pieces.
• (Which is) the very thing it was required to show.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"