Proof: By Euclid
(related to Proposition: 2.10: Sum of Squares (Half))
- 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 $EF$ have been drawn through $E$, parallel to $AD$ [Prop. 1.31], and let $FD$ have been drawn through $D$, parallel to $CE$ [Prop. 1.31].
- And since some straight line $EF$ falls across the parallel straight lines $EC$ and $FD$, the (internal angles) $CEF$ and $EFD$ are thus equal to two right angles [Prop. 1.29].
- Thus, $FEB$ and $EFD$ are less than two right angles.
- And (straight lines) produced from (internal angles whose sum is) less than two right angles meet together [Post. 5] .
- Thus, being produced in the direction of $B$ and $D$, the (straight lines) $EB$ and $FD$ will meet.
- Let them have been produced, and let them meet together at $G$, and let $AG$ have been joined.
- And since $AC$ is equal to $CE$, angle $EAC$ is also equal to (angle) $AEC$ [Prop. 1.5].
- And the (angle) at $C$ (is) a right angle.
- Thus, $EAC$ and $AEC$ [are] each half a right angle [Prop. 1.32].
- So, for the same (reasons), $CEB$ and $EBC$ are also each half a right angle.
- Thus, (angle) $AEB$ is a right angle.
- And since $EBC$ is half a right angle, $DBG$ (is) thus also half a right angle [Prop. 1.15].
- And $BDG$ is also a right angle.
- For it is equal to $DCE$.
- For (they are) alternate (angles) [Prop. 1.29].
- Thus, the remaining (angle) $DGB$ is half a right angle.
- Thus, $DGB$ is equal to $DBG$.
- So side $BD$ is also equal to side $GD$ [Prop. 1.6].
- Again, since $EGF$ is half a right angle, and the (angle) at $F$ (is) a right angle, for it is equal to the opposite (angle) at $C$ [Prop. 1.34], the remaining (angle) $FEG$ is thus half a right angle.
- Thus, angle $EGF$ (is) equal to $FEG$.
- So the side $GF$ is also equal to the side $EF$ [Prop. 1.6].
- And since [$EC$ is equal to $CA$] the square on $EC$ is [also] equal to the square on $CA$.
- Thus, the (sum of the) squares on $EC$ and $CA$ is double the square on $CA$.
- And the (square) on $EA$ is equal to the (sum of the squares) on $EC$ and $CA$ [Prop. 1.47].
- Thus, the square on $EA$ is double the square on $AC$.
- Again, since $FG$ is equal to $EF$, the (square) on $FG$ is also equal to the (square) on $FE$.
- Thus, the (sum of the squares) on $GF$ and $FE$ is double the (square) on $EF$.
- And the (square) on $EG$ is equal to the (sum of the squares) on $GF$ and $FE$ [Prop. 1.47].
- Thus, the (square) on $EG$ is double the (square) on $EF$.
- And $EF$ (is) equal to $CD$ [Prop. 1.34].
- Thus, the square on $EG$ is double the (square) on $CD$.
- But it was also shown that the (square) on $EA$ (is) double the (square) on $AC$.
- Thus, the (sum of the) squares on $AE$ and $EG$ is double the (sum of the) squares on $AC$ and $CD$.
- And the square on $AG$ is equal to the (sum of the) squares on $AE$ and $EG$ [Prop. 1.47].
- Thus, the (square) on $AG$ is double the (sum of the squares) on $AC$ and $CD$.
- And the (sum of the squares) on $AD$ and $DG$ is equal to the (square) on $AG$ [Prop. 1.47].
- Thus, the (sum of the) [squares] on $AD$ and $DG$ is double the (sum of the) [squares] on $AC$ and $CD$.
- And $DG$ (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 in half, and any straight line added to it straight-on, then the sum of the square on the whole (straight line) with the (straight line) having been added, and the (square) on the (straight line) having been added, is double the (sum of the square) on half (the straight line), and the square described on the sum of half (the straight line) and (straight line) having been added, as on one (complete 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"
