Proof: By Euclid

For since parallelogram $DB$ is similar to parallelogram $FB$, they are about the same diagonal [Prop. 6.26].
Let their (common) diagonal $DB$ have been drawn, and let the (rest of the) figure have been described.
Therefore, since (complement) $CF$ is equal to (complement) $FE$ [Prop. 1.43], and (parallelogram) $FB$ is common, the whole (parallelogram) $CH$ is thus equal to the whole (parallelogram) $KE$.
But, (parallelogram) $CH$ is equal to $CG$, since $AC$ (is) also (equal) to $CB$ [Prop. 6.1].
Thus, (parallelogram) $GC$ is also equal to $EK$.
Let (parallelogram) $CF$ have been added to both.
Thus, the whole (parallelogram) $AF$ is equal to the gnomon $LMN$.
Hence, parallelogram $DB$ - that is to say, $AD$ - is greater than parallelogram $AF$.
Thus, for all parallelograms applied to the same straight line, and falling short by a parallelogrammic figure similar, and similarly laid out, to the (parallelogram) described on half (the straight line), the greatest is the [parallelogram] applied to half (the straight line).
(Which is) the very thing it was required to show.
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