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 straight-on to $AG$ [Prop. 1.14].
- So, for the same (reasons), $BA$ is also straight-on 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) $CB$, $BF$, 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. 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 right-angled 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.
Thank you to the contributors under CC BY-SA 4.0!
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"