Proposition 23. Theorem.

565. The sum of any two face-angles of a triedral angle is greater than the third.

Hyp. Let S-ABC be a triedral in which ASC is the greatest face 2.

To prove

ZASB+BSC>ASC.

Proof. In the plane ASC draw SD,

making

Z ASD = ASB.

Draw AC cutting SD in D, take SB = SD, and join AB,

But

SC is common, SB = SD, and BC > DC,

Adding, ZASB+BSC> <ASC.

(120) (Cons.)

Q.E. D.

Proposition 24. Theorem.

566. The sum of the face-angles of any convex polyedral angle is less than four right angles.

Hyp. Let the convex polyedral S be cut by a plane making the section ABCDE a convex polygon.

Proof. From any pt. O within the polygon ABCDE draw OA, OB, OC, E OD, OE.

There are thus formed two sets of

As, one with their common vertex at

S

S, and the other with their common vertex at O, and an equal number of each.

... the sum of the s of these two sets of As is equal. (97)

Because the sum of any two faces of a triedral > the third,

and

.. ¿SAE+ <SAB > <EAB,

ZSBA+ SBC > < ABC, etc.

(565)

Taking the sum of these inequalities, we find that the sum of the s at the bases of the s whose common vertex is S> the sum of the s at the bases of the As whose common vertex is 0.

... the sum of the s at S < the sum of the s at O.

... the sum of the s at S < 4rt. s.

(56) Q.E.D.

Proposition 25. Theorem.

567. Two triedral angles, which have the three faceangles of the one equal respectively to the three face-angles of the other, are either equal or symmetrical.

Hyp. In the triedrals, S and S', let the faces ASB, BSC, CSA = the faces A'S'B', B'S'C', C'S'A'.

To prove S-ABC or symmetrical to S'- A'B'C'.

Proof. In the edges SA, S'A', etc., take the six equal distances SA, SB, SC, S'A', S'B', S'C'; and join AB, BC, CA, A'B', B'C', C'A'.

Then, As SAB, SBC, SCA = As S'A'B', S'B'C', S'C'A'. .. AABC = ▲ A'B'C'.

In the edges SA, S'A', take SD = S'D'.

(108)

In the faces ASB, ASC, and A'S'B', A'S'C', draw DH, DK, and D'H', D'K' to AS and A'S', meeting AB, AC and A'B', A'C', in H, K and H', K'. Join HK and H'K'. Because ADA'D', and DAH =

... rt. ▲ ADH = rt. ▲ A'D'H'.

D'A'H',

... AH = A'H', and DH = D'H'.

(107)

(104)

(536)

In the same way, AK = A'K', and DK = D'K'. .*. ^AHK= ▲ A'H'K', and HK = H'K'. .•. ^HDK= ^ H'D'K', and ▲ HDK = ¿H'D'K'. (108) ... diedral SA = diedral / S'A'. Similarly, diedral s SB, SC = diedral s S'B', S'C'. Now if the equals are arranged in the same order, as in the first two figures, the two triedrals are equal. (557) But if the equals are in reverse order, as in the first and third figures, the triedral s are symmetrical.

(562)

Q.E.D.

EXERCISES.

THEOREMS.

1. If a straight line is parallel to a plane, any plane perpendicular to the line is perpendicular to the plane.

See (503) and (540).

2. If a line is parallel to each of two intersecting planes, it is parallel to their intersection.

3. If a line is perpendicular to a plane, every plane parallel to that line is also perpendicular to the plane.

4. If a line is parallel to each of two planes, the intersections which any plane passing through it makes with the planes are parallel.

5. If a straight line and a plane are perpendicular to the same straight line, they are parallel.

6. If two straight lines are parallel, they are parallel to the common intersection of any two planes passing through them.

7. If the intersections of several planes are parallel, the normals drawn to them from any point are in one plane.

8. If a plane is passed through one of the diagonals of a parallelogram, the perpendiculars to it from the extremities. of the other diagonal are equal.

9. Prove that the sides of an isosceles triangle are equally inclined to any plane through the base.

10. From a point P, PA is drawn perpendicular to a given plane, and from A, AB is drawn perpendicular to a line in that plane: prove that PB is also perpendicular to that line.

11. If the projections of any line upon two intersecting planes are each of them straight lines, prove that the line itself is a straight line.

See (495).

12. If a plane is passed through the middle point of the common perpendicular to two straight lines in space, and parallel to both of these lines, prove that the plane bisects

every straight line which joins a point of one of these lines to a point of the other.

See (526).

13. The projection of a straight line on a plane is a straight line which is either parallel to the first straight line or meets it in the point where it cuts the plane, according as the first line is parallel to the plane or not.

14. From a point P, PA and PB are drawn perpendicular to two planes which intersect in CD, meeting them in A and B; from A, AE is drawn perpendicular to CD: prove that BE is also perpendicular to CD.

15. Two planes which are not parallel are cut by two parallel planes: prove that the intersections of the first two with the last two contain equal angles.

See (519), (525).

16. If two parallel planes be cut by three other planes which have a point, but no line common to all three, and no two of which are parallel, the triangles formed by the intersections of the parallel planes with the other three planes are similar to each other.

See (519), (525).

17. The projections of parallel straight lines on any plane are themselves parallel.

See (509), (525), (519).

18. Two straight lines which intersect are inclined to each other at an angle equal to two-thirds of a right angle, and to a given plane, each, at an angle equal to half a right angle, Prove that their projections on this plane are at right angles to each other.

19. In any triedral angle, the planes bisecting the three diedral angles, all intersect in the same straight line.

See (549).

20. In any triedral angle, the planes which bisect the three. face-angles, and are perpendicular to these faces respectively, all intersect in the same straight line.

From the vertex S, take equal distances SA, SB, SC, on the three edges; the intersections of the three bisecting planes with the plane of ABC are bisecting lines of the sides of ABC, and bave a common intersection (168);.' . etc.