Proof: By Euclid

For let the two straight lines $AB$ and $CD$ be cut by the parallel planes $GH$, $KL$, and $MN$ at the points $A$, $E$, $B$, and $C$, $F$, $D$ (respectively).
I say that as the straight line $AE$ is to $EB$, so $CF$ (is) to $FD$.

fig17e

For let $AC$, $BD$, and $AD$ have been joined, and let $AD$ meet the plane $KL$ at point $O$, and let $EO$ and $OF$ have been joined.
And since two parallel planes $KL$ and $MN$ are cut by the plane $EBDO$, their common sections $EO$ and $BD$ are parallel [Prop. 11.16].
So, for the same (reasons), since two parallel planes $GH$ and $KL$ are cut by the plane $AOFC$, their common sections $AC$ and $OF$ are parallel [Prop. 11.16].
And since the straight line $EO$ has been drawn parallel to one of the sides $BD$ of triangle $ABD$, thus, proportionally, as $AE$ is to $EB$, so $AO$ (is) to $OD$ [Prop. 6.2].
Again, since the straight line $OF$ has been drawn parallel to one of the sides $AC$ of triangle $ADC$, proportionally, as $AO$ is to $OD$, so $CF$ (is) to $FD$ [Prop. 6.2].
And it was also shown that as $AO$ (is) to $OD$, so $AE$ (is) to $EB$.
And thus as $AE$ (is) to $EB$, so $CF$ (is) to $FD$ [Prop. 5.11].
Thus, if two straight lines are cut by parallel planes then they will be cut in the same ratios.
(Which is) the very thing it was required to show.
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