Proof: By Euclid

fig04e

For, in the same figure, let perpendiculars have been conceived (drawn) from (points) $G$ and $H$ to the planes $ABC$ and $DEF$ (respectively).
These clearly turn out to be equal, on account of the pyramids being assumed (to be) of equal height.
And since two straight lines, $GC$ and the perpendicular from $G$, are cut by the parallel planes $ABC$ and $PMN$ they will be cut in the same ratios [Prop. 11.17].
And $GC$ was cut in half by the plane $PMN$ at $N$.
Thus, the perpendicular from $G$ to the plane $ABC$ will also be cut in half by the plane $PMN$.
So, for the same (reasons), the perpendicular from $H$ to the plane $DEF$ will also be cut in half by the plane $STU$.
And the perpendiculars from $G$ and $H$ to the planes $ABC$ and $DEF$ (respectively) are equal.
Thus, the perpendiculars from the triangles $PMN$ and $STU$ to $ABC$ and $DEF$ (respectively, are) also equal.
Thus, the prisms whose bases are triangles $LOC$ and $RVF$, and opposite (sides) $PMN$ and $STU$ (respectively), [are] of equal height.
And, hence, the parallelepiped solids described on the aforementioned prisms [are] of equal height and (are) to one another as their bases [Prop. 11.32].
Likewise, the halves (of the solids) [Prop. 11.28].
Thus, as base $LOC$ is to base $RVF$, so the aforementioned prisms (are) to one another.
(Which is) the very thing it was required to show.
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