Proposition 3

Let there be a pyramid of with the triangular base ABC and vertex D.

I say that the pyramid ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms, and the two prisms are greater than the half of the whole pyramid.

Bisect AB, BC, CA, AD, DB, and DC at the points E, F, G, H, K, and L. Join HE, EG, GH, HK, KL, LH, KF, and FG.

Since AE equals EB, and AH equals DH, therefore EH is parallel to DB. For the same reason HK is also parallel to AB. Therefore HEBK is a parallelogram. Therefore HK equals EB.

But EB equals EA, therefore AE also equals HK.

But AH also equals HD, therefore the two sides EA and AH equal the two sides KH, HD respectively, and the angle EAH equals the angle KHD, therefore the base EH equals the base KD.

Therefore the triangle AEH equals and is similar to the triangle HKD. For the same reason the triangle AHG also equals and is similar to the triangle HLD.

Now, since two straight lines EH and HG meeting one another are parallel to two straight lines KD and DL meeting one another and are not in the same plane, therefore they contain equal angles. Therefore the angle EHG equals the angle KDL.

And, since the two straight lines EH and HG equal the two KD and DL respectively, and the angle EHG equals the angle KDL, therefore the base EG equals the base KL. Therefore the triangle EHG equals and is similar to the triangle KDL. For the same reason the triangle AEG also equals and is similar to the triangle HKL.

Therefore the pyramid with triangular base AEG and vertex H equals and is similar to the pyramid with triangular base HKL and the vertex D.

And, since HK is parallel to AB, one of the sides of the triangle ADB, the triangle ADB is equiangular to the triangle DHK, and they have their sides proportional, therefore the triangle ADB is similar to the triangle DHK. For the same reason the triangle DBC is also similar to the triangle DKL, and the triangle ADC is similar to the triangle DLH.

Now, since the two straight lines BA and AC meeting one another are parallel to the two straight lines KH and HL meeting one another not in the same plane, therefore they contain equal angles. Therefore the angle BAC equals the angle KHL.

And BA is to AC as KH is to HL, therefore the triangle ABC is similar to the triangle HKL.

Therefore the pyramid with the triangular base ABC and vertex D is similar to the pyramid with the triangular base HKL and vertex D.

But the pyramid with the triangular base HKL and vertex D was proved similar to the pyramid with the triangular base AEG and the vertex H. Therefore each of the pyramids AEGH and HKLD is similar to the whole pyramid ABCD.

Next, since BF equals FC, therefore the parallelogram EBFG is double the triangle GFC. And since, if there are two prisms of equal height, and one has a parallelogram as base and the other a triangle, and if the parallelogram is double the triangle, then the prisms are equal. Therefore the prism contained by the two triangles BKF and EHG, and the three parallelograms EBFG, EBKH, and HKFG equals the prism contained by the two triangles GFC and HKL and the three parallelograms KFCL, LCGH, and HKFG.

And it is clear that each of the prisms, namely that with the parallelogram EBFG the base and the straight line HK its opposite, and that with the triangle GFC the base and the triangle HKL its opposite, is greater than each of the pyramids with the triangular bases AEG and HKL and vertices H and D, for, if we join the straight lines EF and EK, the prism with the parallelogram EBFG the base and the straight line HK opposite is greater than the pyramid with the triangular base EBF and vertex K.

But the pyramid with the triangular base EBF and vertex A equals the pyramid with the triangular base AE and the vertex H, for they are contained by equal and similar planes.

Hence the prism with the parallelogram EBF the base and the straight line HK opposite is greater than the pyramid with the triangular base AE and vertex H. But the prism with the parallelogram EBF the base and the straight line HK opposite equals the prism with the triangle GFC the base and the triangle HKL opposite, and the pyramid with the triangular base AEG and vertex H equals the pyramid with the triangular base HKL and vertex D.

Therefore the said two prisms are greater than the said two pyramids with the triangular bases AEG and HKL and vertices H and D. Therefore the whole pyramid with the triangular base ABC and vertex D has been divided into two pyramids equal to one another and into two equal prisms, and the two prisms are greater than the half of the whole pyramid.

Therefore, any pyramid with a triangular base is divided into two pyramids equal and similar to one another, similar to the whole, and having triangular bases, and into two equal prisms, and the two prisms are greater than half of the whole pyramid.