In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

Let $ABC$ be a right-angled triangle having the angle $BAC$ right;I say that the square on $BC$ is equal to the squares on $BA$, $AC$.For let there be described on $BC$ the square $BDEC$,and on $BA$, $AC$ the squares $GB$, $HC$; [Prop. 1.46]through $A$ let $AL$ be drawn parallel to either $BD$ or $CE$, and let $AD$, $FC$ be joined.Then, since each of the angles $BAC$, $BAG$ is right, it follows that with a straight line $BA$, and at the point $A$ on it, the two straight lines $AC$, $AG$ not lying on the same side make the adjacent angles equal to two right angles;therefore $CA$ is in a straight line with $AG$. [Prop. 1.14]For the same reason$BA$ is also in a straight line with $AH$.And, since the angle $DBC$ is equal to the angle $FBA$: for each is right:let the angle $ABC$ be added to each;therefore the whole angle $DBA$ is equal to the whole angle $FBC$. [C.N. 2]And, since $DB$ is equal to $BC$, and $FB$ to $BA$,the two sides $AB$, $BD$ are equal to the two sides $FB$, $BC$ respectively,and the angle $ABD$ is equal to the angle $FBC$;therefore the base $AD$ is equal to the base $FC$,and the triangle $ABD$ is equal to the triangle $FBC$. [Prop. 1.4]Now the parallelogram $BL$ is double of the triangle $ABD$ for they have the same base $BD$ and are in the same parallels $BD$, $AL$. [Prop. 1.41]And the square $GB$ is double of the triangle $FBC$, for they again have the same base $FB$ and are in the same parallels $FB$, $GC$. [Prop. 1.41][But the doubles of equals are equal to one another.]Therefore the parallelogram $BL$ is also equal to the square $GB$.Similarly, if $AE$, $BK$ be joined,the parallelogram $CL$ can also be proved equal to the square $HC$;therefore the whole square $BDEC$ is equal to the two squares $GB$, $HC$. [C.N. 2]And the square $BDEC$ is described on $BC$,and the squares $GB$, $HC$, on $BA$, $AC$.Therefore the square on the side $BC$ is equal to the squares on the sides $BA$, $AC$.Therefore etc.Q.E.D.