let DB, DC be joined.

Now, since DA is equal to AE,

and AB is common,

two sides are equal to two sides;

and the angle DAB is equal to the angle BAE; therefore the base DB is equal to the base BE.

And, since the two sides BD, DC are greater than

and of these DB was proved equal to BE, therefore the remainder DC is greater than the remainde

Now, since DA is equal to AE,

and AC is common,

and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EA

But the angle DAB was also proved equal to the BAE;

therefore the angles DAB, DAC are greater than the BAC.

Similarly we can prove that the remaining angles taken together two and two, are greater than the rem

one.

Therefore etc.

Q. E.

After excluding the obvious case in which all three angles ar Euclid goes on to say "If not, let the angle BAC be greater," withou greater than what. Heiberg is clearly right in saying that he means than BAD, i.e. greater than one of the adjacent angles. This is pr the words at the end "Similarly we can prove," etc. Euclid thus as obvious the case where one of the three angles is not greater than the other two, but proves the remaining cases. This is scientific, but 1 further have excluded as obvious the case in which one angle is grea one of the others but equal to or less than the remaining one.

It would be better, as indicated by Legenare aniu Naustigun by saying that, "If one of the three angles is either equal to or less tha of the other two, it is evident that the sum of those two is greater first. It is therefore only necessary to prove, for the case in which one greater than each of the others, that the sum of the two latter is grea the former.

Accordingly let BAC be greater than each of the other angles." proceed as in Euclid.

PROPOSITION 21.

Any solid angle is contained by plane angles less tha right angles.

Let the angle at A be a solid angle contained by the angles BAC, CAD, DAB;

I say that the angles BAC, CAD,
DAB are less than four right angles.

For let points B, C, D be taken at random on the straight lines AB, AC, AD respectively,

and let BC, CD, DB be joined.

B

Now, since the solid angle at B is contained by the plane angles CBA, ABD, CBD,

any two are greater than the remaining one;

therefore the angles CBA, ABD are greater than the CBD.

For the same reason

the angles BCA, ACD are also greater than the angle and the angles CDA, ADB are greater than the angle therefore the six angles CBA, ABD, BCA, ACD, ADB are greater than the three angles CBD, BCD,

But the three angles CBD, BDC, BCD are equa right angles;

therefore the six angles CBA, ABD, BCA, ACD ADB are greater than two right angles.

and of them the six angles ABC, BCA, ACD, CDA, A DBA are greater than two right angles;

therefore the remaining three angles BAC, CAD, containing the solid angle are less than four right angles. Therefore etc.

Q. E. D

It will be observed that, although Euclid enunciates this propositi any solid angle, he only proves it for the particular case of a trihedral This is in accordance with his manner of proving one case and leavi others to the reader. The omission of the convex polyhedral angl corresponds to the omission, after 1. 32, of the proposition about the i angles of a convex polygon given by Proclus and in most books. The of the present proposition for any convex polyhedral angle can of cou arranged so as not to assume the proposition that the interior angle convex polygon together with four right angles are equal to twice as right angles as the figure has sides.

Let there be any convex polyhedral angle with V as vertex, and le 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; [1. 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.

A

E

Also the sum of the angles of the triangles VAB, VBC, etc., with vertex V are equal to twice as many right angles polygon has sides;

and all the said angles are equal to the sum of (1) the plane angle forming the polyhedral angle and (2) the base angles of the triangle vertex V.

This latter sum is therefore equal to the sum of (3) all the round O and (4) all the interior angles of the polygon.

Now, by Euclid's proposition, of the three angles forming the solid a A, the angles VAE, VAB are together greater than the angle EAB. Similarly, at B, the angles VBA, VBC are together greater than th

ABC.

And so on.

Therefore, by addition, the base angles of the triangles with ve

XI. 21]

PROPOSITION 21

311

[(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.

I. 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.

2.

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.

3. 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.

312

BOOK XI

[XI. 21, 22 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'EF, 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.

in

PROPOSITION 22.

If there be three plane angles of which two, taken together any manner, are greater than the remaining one, and they are contained by equal straight lines, it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines.

Let there be three plane angles ABC, DEF, GHK, of