For let AH be produced in each direction, let any number of straight lines whatever, AK, KL, be made equal to AE, and any number whatever, HM, MN, equal to EH; and let the parallelograms LP, KV, HW, MS and the solids LQ, KR, DM, MT be completed.

Then, since the straight lines LK, KA, AE are equal to one another, the parallelograms LP, KV, AF are also equal to one another, KO, KB, AG are equal to one another, and further LX, KQ, AR are equal to one another, for they are opposite. [[XI. 24](elem.11.24)]

For the same reason the parallelograms EC, HW, MS are also equal to one another, HG, HI, IN are equal to one another, and further DH, MY, NT are equal to one another.

Therefore in the solids LQ, KR, AU three planes are equal to three planes.

But the three planes are equal to the three opposite; therefore the three solids LQ, KR, AU are equal to one another.

For the same reason the three solids ED, DM, MT are also equal to one another.

Therefore, whatever multiple the base LF is of the base AF, the same multiple also is the solid LU of the solid AU.

For the same reason, whatever multiple the base NF is of the base FH, the same multiple also is the solid NU of the solid HU.

And, if the base LF is equal to the base NF, the solid LU is also equal to the solid NU; if the base LF exceeds the base NF, the solid LU also exceeds the solid NU; and, if one falls short, the other falls short.

Therefore, there being four magnitudes, the two bases AF, FH, and the two solids AU, UH, equimultiples have been taken of the base AF and the solid AU, namely the base LF and the solid LU, and equimultiples of the base HF and the solid HU, namely the base NF and the solid NU, and it has been proved that, if the base LF exceeds the base FN, the solid LU also exceeds the solid NU, if the bases are equal, the solids are equal, and if the base falls short, the solid falls short.

Therefore, as the base AF is to the base FH, so is the solid AU to the solid UH. [[V. Def. 5](elem.5.def.5)] Q. E. D.