Kitap 13
17
Let ABCD, CBEF, two planes of the aforesaid cube at right angles to one another, be set out, let the sides AB, BC, CD, DA, EF, EB, FC be bisected at G, H, K, L, M, N, O respectively, let GK, HL, MH, NO be joined, let the straight lines NP, PO, HQ be cut in extreme and mean ratio at the points R, S, T respectively, and let RP, PS, TQ be their greater segments; from the points R, S, T let RU, SV, TW be set up at right angles to the planes of the cube towards the outside of the cube, let them be made equal to RP, PS, TQ, and let UB, BW, WC, CV, VU be joined.
17
I say that the pentagon UBWCV is equilateral, and in one plane, and is further equiangular.
17
For let RB, SB, VB be joined.
17
Then, since the straight line NP has been cut in extreme and mean ratio at R, and RP is the greater segment, therefore the squares on PN, NR are triple of the square on RP. [XIII. 4]
17
But PN is equal to NB, and PR to RU; therefore the squares on BN, NR are triple of the square on RU.
17
But the square on BR is equal to the squares on BN, NR; [I. 47] therefore the square on BR is triple of the square on RU; hence the squares on BR, RU are quadruple of the square on RU.
17
But the square on BU is equal to the squares on BR, RU; therefore the square on BU is quadruple of the square on RU; therefore BU is double of RU.
17
But VU is also double of UR, inasmuch as SR is also double of PR, that is, of RU; therefore BU is equal to UV.
