Proof: By Euclid
(related to Proposition: Prop. 12.06: Sizes of Pyramids of Same Height with Polygonal Bases are as Bases)
 Let there be pyramids of the same height whose bases (are) the polygons $ABCDE$ and $FGHKL$, and apexes the points $M$ and $N$ (respectively).
 I say that as base $ABCDE$ is to base $FGHKL$, so pyramid $ABCDEM$ (is) to pyramid $FGHKLN$.
 For let $AC$, $AD$, $FH$, and $FK$ have been joined.
 Therefore, since $ABCM$ and $ACDM$ are two pyramids having triangular bases and equal height, they are to one another as their bases [Prop. 12.5].
 Thus, as base $ABC$ is to base $ACD$, so pyramid $ABCM$ (is) to pyramid $ACDM$.
 And, via composition, as base $ABCD$ (is) to base $ACD$, so pyramid $ABCDM$ (is) to pyramid $ACDM$ [Prop. 5.18].
 But, as base $ACD$ (is) to base $ADE$, so pyramid $ACDM$ (is) also to pyramid $ADEM$ [Prop. 12.5].
 Thus, via equality, as base $ABCD$ (is) to base $ADE$, so pyramid $ABCDM$ (is) to pyramid $ADEM$ [Prop. 5.22].
 And, again, via composition, as base $ABCDE$ (is) to base $ADE$, so pyramid $ABCDEM$ (is) to pyramid $ADEM$ [Prop. 5.18].
 So, similarly, it can also be shown that as base $FGHKL$ (is) to base $FGH$, so pyramid $FGHKLN$ (is) also to pyramid $FGHN$.
 And since $ADEM$ and $FGHN$ are two pyramids having triangular bases and equal height, thus as base $ADE$ (is) to base $FGH$, so pyramid $ADEM$ (is) to pyramid $FGHN$ [Prop. 12.5].
 But, as base $ADE$ (is) to base $ABCDE$, so pyramid $ADEM$ (was) to pyramid $ABCDEM$.
 Thus, via equality, as base $ABCDE$ (is) to base $FGH$, so pyramid $ABCDEM$ (is) also to pyramid $FGHN$ [Prop. 5.22].
 But, furthermore, as base $FGH$ (is) to base $FGHKL$, so pyramid $FGHN$ was also to pyramid $FGHKLN$.
 Thus, via equality, as base $ABCDE$ (is) to base $FGHKL$, so pyramid $ABCDEM$ (is) also to pyramid $FGHKLN$ [Prop. 5.22].
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"