G.549 sides opposite planes — Heath

XI.38

If the sides of the opposite planes of a cube be bisected, and planes be carried through the points of section, the common section of the planes and the diameter of the cube bisect one another.

For let the sides of the opposite planes < C F >, < A H > of the cube < A F > be bisected at the points K, L, M, N, O, Q, P, R, and through the points of section let the planes < K N >, < O R > be carried; let / U S / be the common section of the planes, and / D G / the diameter of the cube < A F >.

I say that / U T / is equal to / T S /, and / D T / to / T G /.

For let / D U /, / U E /, / B S /, / S G / be joined.

Then, since / D O / is parallel to / P E /,

the alternate angles < D O U ), < U P E ) are equal to one another. [G.56]

And, since / D O / is equal to / P E /, and / O U / to / U P /, and they contain equal angles,

therefore the base / D U / is equal to the base / U E /, the triangle < D O U > is equal to the triangle < P U E >, and the remaining angles are equal to the remaining angles; [G.24] therefore the angle < O U D ) is equal to the angle < P U E ).

[G.39]For this reason / D U E / is a straight line.

For the same reason, / B S G / is also a straight line, and / B S / is equal to / S G /.

Now, since / C A / is equal and parallel to / D B /, while / C A / is also equal and parallel to / E G /,

[G.511]therefore / D B / is also equal and parallel to / E G /.

And the straight lines / D E /, / B G / join their extremities ;

[G.60]therefore / D E / is parallel to / B G /.

Therefore the angle < E D T ) is equal to the angle < B G T ), for they are alternate ; [G.56] and the angle < D T U ) is equal to the angle < G T S ). [G.40]

Therefore < D T U >, < G T S > are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is, / D U / equal to / G S /, for they are the halves of / D E /, / B G / ; therefore they will also have the remaining sides equal to the remaining sides. [G.51]

Therefore / D T / is equal to / T G /, and / U T / to / T S /.

Therefore etc. Q. E. D.

Dependency Graph

Required for

G.600, G.601