Book 13
17
Similarly it can be proved that each of the straight lines BW, WC, CV is also equal to each of the straight lines BU, UV.
17
Therefore the pentagon BUVCW is equilateral.
17
I say next that it is also in one plane.
17
For let PX be drawn from P parallel to each of the straight lines RU, SV and towards the outside of the cube, and let XH, HW be joined; I say that XHW is a straight line.
17
For, since HQ has been cut in extreme and mean ratio at T, and QT is its greater segment, therefore, as HQ is to QT, so is QT to TH.
17
But HQ is equal to HP, and QT to each of the straight lines TW, PX; therefore, as HP is to PX, so is WT to TH.
17
And HP is parallel to TW, for each of them is at right angles to the plane BD; [XI. 6] and TH is parallel to PX, for each of them is at right angles to the plane BF. [id.]
17
But if two triangles, as XPH, HTW, which have two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining straight lines will be in a straight line; [VI. 32] therefore XH is in a straight line with HW.
17
But every straight line is in one plane; [XI. 1] therefore the pentagon UBWCV is in one plane.
17
I say next that it is also equiangular.
17
For, since the straight line NP has been cut in extreme and mean ratio at R, and PR is the greater segment, while PR is equal to PS, therefore NS has also been cut in extreme and mean ratio at P, and NP is the greater segment; [XIII. 5] therefore the squares on NS, SP are triple of the square on NP. [XIII. 4]
17
But NP is equal to NB, and PS to SV; therefore the squares on NS, SV are triple of the square on NB; hence the squares on VS, SN, NB are quadruple of the square on NB.
17
But the square on SB is equal to the squares on SN, NB; therefore the squares on BS, SV, that is, the square on BV —for the angle VSB is right—is quadruple of the square on NB; therefore VB is double of BN.
17
But BC is also double of BN; therefore BV is equal to BC.
