Commentary on Proposition XI.38¶
Back to Proposition XI.38.
Euclid enunciates this proposition of a cube only, though it is true of any parallelepiped, no doubt because its truth for a cube is all that was wanted for the only proposition where it is needed, viz. XIII. 17.
Simson remarks that it should be proved that the straight lines bisecting the corresponding opposite sides of opposite planes are in one plane. This is, however, clear because e.g. since DK, CL are equal and parallel, KL is equal and parallel to CD. And, since KL, AB are both parallel to DC, KL is parallel to AB. And lastly, since KL, MN are both parallel to AB, KL is parallel to MN and therefore in one plane with it.
The essential thing to be proved is that the plane passing through the opposite edges DB, EG passes through the straight line US, since, only if this be the case, can US, DG intersect one another.
To prove this we have only to prove that, if DU, UE and BS, SG be joined, DUE and BSG are both straight lines.
Now, since DO is parallel to PE,
$$\angle DOU = \angle EPU.$$
Thus, in the $\triangle\text{s}$ DUO, EUP, two sides DO, OU are equal to two sides EP, PU, and the included angles are equal.
Therefore
$$\triangle DUO \equiv \triangle EUP,$$
$$DU = UE,$$
and
$$\angle DUO = \angle EUP,$$
so that DUE is a straight line, bisected at U. Similarly BSG is a straight line, bisected at S.
Thus the plane through DB, EG (DB, EG being equal and parallel) contains the straight lines DUE, BSG (which are therefore equal and parallel also) and also [XI. 7] the straight lines US, DG (which accordingly intersect).
In $\triangle\text{s}$ DTU, GTS, the angles UDT, SGT are equal (being alternate), and the angles UTD, STG are also equal (being vertically opposite), while DU (half of DE) is equal to GS (half of BG).
Therefore [I. 26] the triangles DTU, GTS are equal in all respects, so that
$$DT = TG,$$
$$UT = TS.$$