Proof: By Euclid
(related to Proposition: 1.47: Pythagorean Theorem)
 For let the square $BDEC$ have been described on $BC$, and (the squares) $GB$ and $HC$ on $AB$ and $AC$ (respectively) [Prop. 1.46].
 And let $AL$ have been drawn through point $A$ parallel to either of $BD$ or $CE$ [Prop. 1.31].
 And let $AD$ and $FC$ have been joined.
 And since angles $BAC$ and $BAG$ are each right angles, then two straight lines $AC$ and $AG$, not lying on the same side, make the adjacent angles with some straight line $BA$, at the point $A$ on it, (whose sum is) equal to two right angles.
 Thus, $CA$ is straighton to $AG$ [Prop. 1.14].
 So, for the same (reasons), $BA$ is also straighton to $AH$.
 And since angle $DBC$ is equal to $FBA$, for (they are) both right angles, let $ABC$ have been added to both.
 Thus, the whole (angle) $DBA$ is equal to the whole (angle) $FBC$.
 And since $DB$ is equal to $BC$, and $FB$ to $BA$, the two (straight lines) $DB$, $BA$ are equal to the two (straight lines) $CB$, $BF$,^{1} respectively.
 And angle $DBA$ (is) equal to angle $FBC$.
 Thus, the base $AD$ [is] equal to the base $FC$, and the triangle $ABD$ is equal to the triangle $FBC$ [Prop. 1.4].
 And parallelogram $BL$ [is] double (the area) of triangle $ABD$.
 For they have the same base, $BD$, and are between the same parallels, $BD$ and $AL$ [Prop. 1.41].
 And square $GB$ is double (the area) of triangle $FBC$.
 For again they have the same base, $FB$, and are between the same parallels, $FB$ and $GC$ [Prop. 1.41].
 And the doubles of equal things are equal to one another.^{2} Thus, the parallelogram $BL$ is also equal to the square $GB$.
 So, similarly, $AE$ and $BK$ being joined, the parallelogram $CL$ can be shown (to be) equal to the square $HC$.
 Thus, the whole square $BDEC$ is equal to the (sum of the) two squares $GB$ and $HC$.
 And the square $BDEC$ is described on $BC$, and the (squares) $GB$ and $HC$ on $BA$ and $AC$ (respectively).
 Thus, the square on the side $BC$ is equal to the (sum of the) squares on the sides $BA$ and $AC$.
 Thus, in rightangled triangles, the square on the side subtending the right angle is equal to the (sum of the) squares on the sides surrounding the right[angle].
 (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"
Footnotes