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¶ The 6. Proposition. The 6. Probleme. In an Octohedron geuen, to inscribe a trilater equilater Pyramis.
SVppose tha•• the Octohedron where•••• the Tetrahedron is required to be ins••ri••••••, be ABGDEI. Take 〈…〉〈…〉 bases of the Octo••••dron, that is,* 1.1 〈…〉〈…〉 close in the lowe•••• triangle BGD, namely, AE••, HED, IGD: and let the fourth be AIB, which is opposite to the lowest trian∣gle before put, namely, to EGD. And take the centres of those fower bases, which let be the pointes H, C, N, ••. And vpon the triangle HCN erecte a pyramis HCNL. Now ••orasmuch as these two bases of the Octohedron, namely, AGE and ABI are set vpon the right lines EG and BI which are opposite the one to the other•• in the square GEBI of the Octohedron, from the poin•• A dra••e by the centres of the bases, namely, by the centres H, L, perpendicular lines AHF, ALK, cutting the lines EG and BI 〈◊〉〈◊〉 two equall partes in the pointes F, K (by the Corollary of the 1•• of the thirtenth). Wherfore
A Corollary.
The bases of a Pyramis inscribed in an Octohedron, are parallels to the bases of the Octohedron. For forasmuch as the sides of the bases of the Pyramis touching the one the other, are parallels to the sides of the Octohedron which also touch the one the other, as for example, HL was proued to be a parallel to GI, and LC to DI, therefore, by the 15. of the eleuenth, the plaine superficies which is drawen by the lines HL and LC, is a parallel to the plaine superficies drawen by the lines GI and DI. And so likewise of the rest.
Second Corollary.
A right line ioyning together the centres of the opposite bases of the Octo∣hedron, is sesquialter to the perpendicular line drawen from the angle of the in∣scribed