Commentary on Proposition XII.6

Back to Proposition XII.6.

[Folio 0397 | p. 393]

It will be seen that, in order to obtain the proportion (base ABCDE) $: \triangle$ ADE $=$ (pyramid MABCDE) $:$ (pyramid MADE), Euclid employs V. 18 (componendo) twice over, with an ex aequali step [V. 22] intervening.

We might arrive at it more concisely by using V. 24 extended to any number of antecedents.

Thus

$\triangle$ ABC $: \triangle$ ADE $=$ (pyramid MABC) $:$ (pyramid MADE), $\triangle$ ACD $: \triangle$ ADE $=$ (pyramid MACD) $:$ (pyramid MADE),

and lastly

$\triangle$ ADE $: \triangle$ ADE $=$ (pyramid MADE) $:$ (pyramid MADE).

Therefore, adding the antecedents [V. 24], we have

(polygon ABCDE) $: \triangle$ ADE $=$ (pyramid MABCDE) $:$ (pyramid MADE).

Again, since the pyramids MADE, NFGH are of the same height,

$\triangle$ ADE $: \triangle$ FGH $=$ (pyramid MADE) $:$ (pyramid NFGH).

Lastly, using the same argument for the pyramid NFGHKL as for MABCDE, and inverting, we have

$\triangle$ FGH $:$ (polygon FGHKL) $=$ (pyramid NFGH) $:$ (pyramid NFGHKL).

Thus from the three proportions, ex aequali,

(polygon ABCDE) $:$ (polygon FGHKL) $=$ (pyramid MABCDE) $:$ (pyramid NFGHKL).