# Proof: By Euclid

• For since $ABCD$ is a parallelogram, $AD$ is equal to $BC$ [Prop. 1.34].
• So, for the same (reasons), $EF$ is also equal to $BC$.
• So $AD$ is also equal to $EF$.
• And $DE$ is common.
• Thus, the whole (straight line) $AE$ is equal to the whole (straight line) $DF$.
• And $AB$ is also equal to $DC$.
• So the two (straight lines) $EA$, $AB$ are equal to the two (straight lines) $FD$, $DC$, respectively.
• And angle $FDC$ is equal to angle $EAB$, the external to the internal [Prop. 1.29].
• Thus, the base $EB$ is equal to the base $FC$, and triangle $EAB$ will be equal to triangle $DFC$ [Prop. 1.4].
• Let $DGE$ have been taken away from both.
• Thus, the remaining trapezium $ABGD$ is equal to the remaining trapezium $EGCF$.
• Let triangle $GBC$ have been added to both.
• Thus, the whole parallelogram $ABCD$ is equal to the whole parallelogram $EBCF$.
• Thus, parallelograms which are on the same base and between the same parallels are equal to one another.
• (Which is) the very thing it was required to show.

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