# Proof: By Euclid

• For let the (straight line) $BD$ have been produced in each direction to points $H$ and $L$, and let [any number] (of straight lines) $BG$ and $GH$ be made equal to base $BC$, and any number (of straight lines) $DK$ and $KL$ equal to base $CD$.
• And let $AG$, $AH$, $AK$, and $AL$ have been joined.
• And since $CB$, $BG$, and $GH$ are equal to one another, triangles $AHG$, $AGB$, and $ABC$ are also equal to one another [Prop. 1.38].
• Thus, as many times as base $HC$ is (divisible by) base $BC$, so many times is triangle $AHC$ also (divisible) by triangle $ABC$.
• So, for the same (reasons), as many times as base $LC$ is (divisible) by base $CD$, so many times is triangle $ALC$ also (divisible) by triangle $ACD$.
• And if base $HC$ is equal to base $CL$ then triangle $AHC$ is also equal to triangle $ACL$ [Prop. 1.38].
• And if base $HC$ exceeds base $CL$ then triangle $AHC$ also exceeds triangle $ACL$.1
• And if ($HC$ is) less (than $CL$ then $AHC$ is also) less (than $ACL$).
• So, their being four magnitudes, two bases, $BC$ and $CD$, and two triangles, $ABC$ and $ACD$, equal multiples have been taken of base $BC$ and triangle $ABC$ - (namely), base $HC$ and triangle $AHC$ - and other random equal multiples of base $CD$ and triangle $ADC$ - (namely), base $LC$ and triangle $ALC$.
• And it has been shown that if base $HC$ exceeds base $CL$ then triangle $AHC$ also exceeds triangle $ALC$, and if ($HC$ is) equal (to $CL$ then $AHC$ is also) equal (to $ALC$), and if ($HC$ is) less (than $CL$ then $AHC$ is also) less (than $ALC$).
• Thus, as base $BC$ is to base $CD$, so triangle $ABC$ (is) to triangle $ACD$ [Def. 5.5] .
• And since parallelogram $EC$ is double triangle $ABC$, and parallelogram $FC$ is double triangle $ACD$ [Prop. 1.34], and parts have the same ratio as similar multiples [Prop. 5.15], thus as triangle $ABC$ is to triangle $ACD$, so parallelogram $EC$ (is) to parallelogram $FC$.
• In fact, since it was shown that as base $BC$ (is) to $CD$, so triangle $ABC$ (is) to triangle $ACD$, and as triangle $ABC$ (is) to triangle $ACD$, so parallelogram $EC$ (is) to parallelogram $CF$, thus, also, as base $BC$ (is) to base $CD$, so parallelogram $EC$ (is) to parallelogram $FC$ [Prop. 5.11].
• Thus, triangles and parallelograms which are of the same height are to one another as their bases.
• (Which is) the very thing it was required to show.

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