Commentary on Proposition XII.4

Back to Proposition XII.4.

LEMMA.

But that, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite, we must prove as follows.

For in the same figure let perpendiculars be conceived drawn from G, H to the planes ABC, DEF; these are of course equal because, by hypothesis, the pyramids are of equal height.

Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC, PMN,
[XI. 17]they will be cut in the same ratios.
And GC is bisected by the plane PMN at N;
therefore the perpendicular from G to the plane ABC will also be bisected by the plane PMN.
For the same reason the perpendicular from H to the plane DEF will also be bisected by the plane STU.
And the perpendiculars from G, H to the planes ABC, DEF are equal;
therefore the perpendiculars from the triangles PMN, STU to the planes ABC, DEF are also equal.
Therefore the prisms in which the triangles LOC, RVF are bases, and PMN, STU their opposites, are of equal height.
Hence also the parallelepipedal solids described from the said prisms are of equal height and are to one another as their bases; [XI. 32] therefore their halves, namely the said prisms, are to one another as the base LOC is to the base RVF. Q. E. D.

We can incorporate the lemma at the end of the proposition and summarise the proof thus.

Since LO is parallel to AB, $\triangle\text{s}$ ABC, LOC are similar. In like manner $\triangle\text{s}$ DEF, RVF are similar. And, since BC $:$ CO $=$ EF $:$ FV, $\triangle$ ABC $: \triangle$ LOC $= \triangle$ DEF $: \triangle$ RVF, [VI. 22] and, alternately, $\triangle$ ABC $: \triangle$ DEF $= \triangle$ LOC $: \triangle$ RVF.

Now the prisms (LOC, PMN) and (RVF, STU) are equal in height: for the perpendiculars from G, H on the bases ABC, DEF are divided by the planes PMN, STU (parallel to the bases) in the same proportion as GC, HF are divided by those planes [XI. 17], i.e. they are bisected; hence the heights of the prisms, being half the equal heights of the pyramids, are equal.

And the prisms are the halves respectively of parallelepipeds of the same height on parallelogrammic bases double of the $\triangle\text{s}$ LOC, RVF respectively; [XI. 28 and note] hence they are in the same ratio as those parallelepipeds, and therefore as their bases [XI. 32].

Therefore $(\text{prism }$ LOC, PMN $) : (\text{prism }$ RVF, STU $) = \triangle$ LOC $: \triangle$ RVF $= \triangle$ ABC $: \triangle$ DEF.

And since the other prisms in the pyramids are equal to these prisms respectively, $(\text{sum of prisms in }$ GABC $) : (\text{sum of prisms in }$ HDEF $) = \triangle$ ABC $: \triangle$ DEF.

Similarly, if the pyramids GPMN, HSTU be divided in like manner, and also the pyramids PAKL, SDQR, we shall have e.g. $(\text{sum of prisms in }$ GPMN $) : (\text{sum of prisms in }$ HSTU $) = \triangle$ PMN $: \triangle$ STU $= \triangle$ ABC $: \triangle$ DEF, and similarly for the second pair of pyramids.

The process may be continued indefinitely, and we shall always have $(\text{sum of prisms in }$ GABC $) : (\text{sum of prisms in }$ HDEF $) = \triangle$ ABC $: \triangle$ DEF.