Commentary on Proposition XI.21¶
Back to Proposition XI.21.
It will be observed that, although Euclid enunciates this proposition for any solid angle, he only proves it for the particular case of a trihedral angle. This is in accordance with his manner of proving one case and leaving the others to the reader. The omission of the convex polyhedral angle here corresponds to the omission, after I. 32, of the proposition about the interior angles of a convex polygon given by Proclus and in most books. The proof of the present proposition for any convex polyhedral angle can of course be arranged so as not to assume the proposition that the interior angles of a convex polygon together with four right angles are equal to twice as many right angles as the figure has sides.
Let there be any convex polyhedral angle with V as vertex, and let it be cut by any plane meeting its faces in, say, the polygon ABCDE.
Take O any point within the polygon, and in its plane, and join OA, OB, OC, OD, OE.
Then all the angles of the triangles with vertex O are equal to twice as many right angles as the polygon has sides; [I. 32] therefore the interior angles of the polygon together with all the angles round O are equal to twice as many right angles as the polygon has sides.
Also the sum of the angles of the triangles VAB, VBC, etc., with vertex V are equal to twice as many right angles as the polygon has sides; and all the said angles are equal to the sum of (1) the plane angles at V forming the polyhedral angle and (2) the base angles of the triangles with vertex V.
This latter sum is therefore equal to the sum of (3) all the angles round O and (4) all the interior angles of the polygon.
Now, by Euclid’s proposition, of the three angles forming the solid angle at A, the angles VAE, VAB are together greater than the angle EAB.
Similarly, at B, the angles VBA, VBC are together greater than the angle ABC.
And so on.
Therefore, by addition, the base angles of the triangles with vertex V [(2) above] are together greater than the sum of the angles of the polygon [(4) above].
Hence, by way of compensation, the sum of the plane angles at V [(1) above] is less than the sum of the angles round O [(3) above].
But the latter sum is equal to four right angles; therefore the plane angles forming the polyhedral angle are together less than four right angles.
The proposition is only true of convex polyhedral angles, i.e. those in which the plane of any face cannot, if produced, ever cut the solid angle.
There are certain propositions relating to equal (and symmetrical) trihedral angles which are necessary to the consideration of the polyhedra dealt with by Euclid, all of which (as before remarked) have trihedral angles only.
Two trihedral angles are equal if two face angles and the included dihedral angle of the one are respectively equal to two face angles and the included dihedral angle of the other, the equal parts being arranged in the same order.
Two trihedral angles are equal if two dihedral angles and the included face angle of the one are respectively equal to two dihedral angles and the included face angle of the other, all equal parts being arranged in the same order.
These propositions are proved immediately by superposition.
Two trihedral angles are equal if the three face angles of the one are respectively equal to the three face angles of the other, and all are arranged in the same order.
Let V—ABC and V'—A'B'C' be two trihedral angles such that the angle AVB is equal to the angle A'V'B', the angle BVC to the angle B'V'C', and the angle CVA to the angle C'V'A'.
We first prove that corresponding pairs of face angles include equal dihedral angles.
E.g., the dihedral angle formed by the plane angles CVA, AVB is equal to that formed by the plane angles C'V'A', A'V'B'.
Take points A, B, C on VA, VB, VC and points A', B', C' on V'A', V'B', V'C', such that VA, VB, VC, V'A', V'B', V'C' are all equal.
Join BC, CA, AB, B'C', C'A', A'B'.
Take any point D on AV, and measure A'D' along A'V' equal to AD.
From D draw DE in the plane AVB, and DF in the plane CVA, perpendicular to AV. Then DE, DF will meet AB, AC respectively, the angles VAB, VAC, the base angles of two isosceles triangles, being less than right angles.
Join EF.
Draw the triangle D'E'F' in the same way.
Now, by means of the hypothesis and construction, it appears that the triangles VAB, V'A'B' are equal in all respects. So are the triangles VAC, V'A'C', and the triangles VBC, V'B'C'. Thus BC, CA, AB are respectively equal to B'C', C'A', A'B', and the triangles ABC, A'B'C' are equal in all respects.
Now, in the triangles ADE, A'D'E', the angles ADE, DAE are equal to the angles A'D'E', D'A'E' respectively, and AD is equal to A'D'. Therefore the triangles ADE, A'D'E' are equal in all respects. Similarly the triangles ADF, A'D'F' are equal in all respects.
Thus, in the triangles AEF, A'E'F', EA, AF are respectively equal to E'A', A'F', and the angle EAF is equal to the angle E'A'F' (from above); therefore the triangles AEF, A'E'F' are equal in all respects.
Lastly, in the triangles DEF, D'E'F', the three sides are respectively equal to the three sides; therefore the triangles are equal in all respects. Therefore the angles EDF, E'D'F' are equal.
But these angles are the measures of the dihedral angles formed by the planes CVA, AVB and by the planes C'V'A', A'V'B' respectively. Therefore these dihedral angles are equal. Similarly for the other two dihedral angles. Hence the trihedral angles coincide if one is applied to the other; that is, they are equal.
To understand what is implied by “taken in the same order” we may suppose ourselves to be placed at the vertices, and to take the faces in clockwise direction, or the reverse, for both angles.
If the face angles and dihedral angles are taken in reverse directions, i.e. in clockwise direction in one and in counterclockwise direction in the other, then, if the other conditions in the above three propositions are fulfilled, the trihedral angles are not equal but symmetrical.
If the faces of a trihedral angle be produced beyond the vertex, they form another trihedral angle. It is easily seen that these vertical trihedral angles are symmetrical.