Commentary on Proposition XI.35¶
Back to Proposition XI.35.
This proposition is required for the next, where it is necessary to know that, if in two equiangular parallelepipeds equal angles, one in each, be contained by three plane angles respectively, one of which is an angle of the parallelogram forming the base in one parallelepiped, while its equal is likewise in the base of the other, and the edges in which the two remaining angles forming the solid angles meet are equal, the parallelepipeds are of the same height.
Bearing in mind the definition of the inclination of a straight line to a plane, we might enunciate the proposition more shortly thus.
If there be two trihedral angles identically equal to one another, corresponding edges in each are equally inclined to the planes through the other two edges respectively.
The proof, which is necessarily somewhat long, may be summarised thus. It is required to prove that the angles GAL, MDN in the figure are equal, G, M being any points on AG, DM, and GL, MN perpendicular to the planes BAC, EDF respectively.
If AH is made equal to DM, and HK is drawn in the plane GAL parallel to GL,
$$\text{\it HK\rm{} is also perpendicular to the plane \it BAC\rm{}.} \qquad \text{[XI. 8]}$$
Draw KB, KC perpendicular to AB, AC respectively and NE, NF perpendicular to DE, DF respectively, and complete the figures.
Now (1)
$$\left.\begin{aligned} HA^2 &= HK^2 + KA^2 \\ &= HK^2 + KC^2 + CA^2 \\ &= HC^2 + CA^2 \end{aligned}\right\} \qquad \text{[I. 47]}$$
$$\begin{aligned} \text{Therefore}\qquad \angle HCA &= \text{a right angle.} \\ \text{Similarly}\qquad \angle MFD &= \text{a right angle.} \end{aligned}$$
$\Delta$s HAC, MDF have therefore two angles equal and one side.
$$\text{Therefore}\qquad \Delta HAC \equiv \Delta MDF\text{, and } AC = DF. \qquad \text{[I. 26]}$$
Similarly $\Delta$ HAB $\equiv \Delta$ MDE, and AB $=$ DE.
Hence $\Delta$s ABC, DEF are equal in all respects, so that BC $=$ EF, and
$$\begin{aligned} \angle ABC &= \angle DEF, \\ \angle ACB &= \angle DFE. \end{aligned}$$
Therefore the complements of these angles are equal, i.e.
$$\begin{aligned} \angle KBC &= \angle NEF, \\ \text{and}\qquad \angle KCB &= \angle NFE. \end{aligned}$$
The $\Delta$s KBC, NEF have two angles equal and one side, and are therefore equal in all respects, so that
$$\begin{aligned} KB &= NE, \\ KC &= NF. \end{aligned}$$
The right-angled triangles KAC, NDF are equal in all respects, since AC $=$ DF [(2) above], KC $=$ NF.
$$\text{Consequently}\qquad AK = DN.$$
In $\Delta$s HAK, MDN,
$$\begin{aligned} HK^2 + KA^2 &= HA^2 \\ &= MD^2\text{, by hypothesis,} \\ &= MN^2 + ND^2. \end{aligned}$$
Subtracting the equals KA ², ND ²,
$$\begin{aligned} \text{we have}\qquad HK^2 &= MN^2, \\ \text{or}\qquad HK &= MN. \end{aligned}$$
$\Delta$s HAK, MDN are now equal in all respects, by I. 8 and I. 4, and therefore
$$\angle HAK = \angle MDN.$$
The Porism is merely a statement of the result arrived at in (8).
Legendre uses, practically, the construction and argument of this proposition to prove the theorem given under (3) of the note on XI. 21 above that In two equal trihedral angles, corresponding pairs of face angles include equal dihedral angles. This fact is readily deduced from the above proposition.
Since [(1)] HC, KC are both perpendicular to AC, and MF, NF both perpendicular to DF, the angles HCK, MFN are the measures of the dihedral angles between the planes HAC, BAC, and MDF, EDF respectively. [XI. Def. 6]
$$\begin{aligned} \text{By (6),}\qquad KC &= NF, \\ \text{and, by (8),}\qquad HK &= MN, \end{aligned}$$
while the angles HKC, MNF, both being right, are equal. Consequently the $\Delta$s HCK, MFN are equal in all respects, [I. 4] so that
$$\angle HCK = \angle MFN.$$
Simson substituted a different proof of (1) in the above summary, as follows.
Since HK is perpendicular to the plane BAC, the plane HBK, passing through HK, is also perpendicular to the plane BAC. [XI. 18]
And AB, being drawn in the plane BAC perpendicular to BK, the common section of the planes HBK, BAC, is perpendicular to the plane HBK [XI. Def. 4], and is therefore perpendicular to every straight line meeting it in that plane [XI. Def. 3].
Hence the angle ABH is a right angle.
I think Euclid’s proof much preferable to this with its references to definitions which are more of the nature of theorems.