# 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) $CB$, $BF$,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 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.

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### References

1. The Greek text has "$FB$, $BC$", which is obviously a mistake (translator's note).