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PROPOSITION XXIV. THEOREM

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

B

Hyp. ZAVC is the greatest face angle of triedral angle V-ABC.

LAVB+L BVC > LAVC.

To prove

Proof.

In the face AVC draw VD equal to VB, making

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Adding the equal angles AVB and AVD,

ZAVB+ ZBVC > ZDVC + LAVD,
LAVB + ZBVC > LAVC.

or

Q.E.D.

Ex. 1063. In the diagram for Prop. XXIV,

LAVD+ZDVB < LAVC +2CDB. * Ex. 1064. In the same diagram, if E is a point in A ABC,

LAVE + ZBVE+ZCVE<LAVB + ZBVC + LCVA, but

> (ZAVB + ZBVC + ZCVA). Ex. 1065. In the diagram for Prop. XXII, if in plane MN AF be drawn so as to make Z FAC obtuse, then ZFAB is obtuse.

PROPOSITION XXV. THEOREM

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

B В

Hyp. V-ABCDE is any convex polyedral angle.

To prove the sum of 1 AVB, BVC, etc., is less than four right angles.

Proof. Construct a plane intersecting the edges in A, B, C, etc., the faces in AB, BC, CD, etc. Then ABCDE is a convex polygon.

Join any point 0 in plane ABC to A, B, C, etc.
Then

Z VBA + Z VBC > ZABC, and

Z VCB + Z VCD > Z BCD, etc. (522) Therefore the sum of all the base angles of the triangles whose vertex is V is greater than the sum of all the base angles of the triangles whose vertex is 0.

But the sum of all the angles of the triangles whose vertex is V is equal to the sum of all the angles of the triangles whose vertex is 0.

(Why ?) Subtracting the base angles from the equal sums, the sum of all the angles at V is less than the sum of all the angles at 0.

But the sum of all the angles at () is equal to four right angles.

Whence, the sum of all the angles at V is less than four right angles.

PROPOSITION XXVI. THEOREM 524. Two triedral angles are equal if two face angles and the included diedral angle of the one are respectively equal to two face angles and the included diedral angle of the other, the equal parts being arranged in the same order.

[Proof by superposition.]

Q.E.D.

PROPOSITION XXVII. THEOREM

525. Two triedral angles are equal if two diedral angles and the included face angle of the one are respectively equal to two diedral angles and the included face angle of the other, all equal parts being arranged in the same order.

[Proof by superposition.]

PROPOSITION XXVIII. THEOREM

526. Two triedral 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.

[blocks in formation]

Hyp. In triedral angles V-ABC and V'-A'B'C",

[blocks in formation]

Draw AB, BC, CA, A'B', B'C', and 'A'.

[blocks in formation]

On AV and A'V', respectively, lay off AD equal to A'D'. Draw DE in face AVB, and DF in face AVC, perpendicular to VA. These lines meet AB and BC, respectively, since triangles AVB and AVC are isosceles.

In like manner, draw D'E' and D'F', and join EF and E'F'.
[To be completed by the student.]
Hint.— Prove

A ABC = A A'B'C'.
AADE = A A'D' E'.
A ADF = A ADF.
Δ ΑΕF =Δ Α'E' F'.
A DEF=A D'E'F'.

1

527. DEF. Vertical polyedral angles are polyedral angles which have a common vertex, and the edges of one are prolongations of the edges of the other.

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PROPOSITION XXIX 528. Two triedral angles are symmetrical :

(1) If two face angles and the included diedral angle of the one are respectively equal to two face angles and the included diedral angle of the other, or, (2) If two diedral angles and the included face angle of the one are respectively equal to two diedral angles and the included face angle of the other, or, (3) Three face angles of the one are equal respectively to three face angles of the other, provided all equal parts are arranged in reverse order.

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Proof. If A and B are the given triedral angles, construct triedral angle C symmetrical to A.

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