In right-angled triangles,  the square on the side subtending the right-angle is equal to the (sum of the) squares on the sides containing the right-angle.  Let $ABC$ be a right-angled triangle having the angle $BAC$a right-angle. I say that the square on $BC$ is equal to the (sum of the) squares on $BA$ and $AC$.  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.