Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

PROPOSITION XXV. THEOREM. 489. An isosceles trihedral angle and its symmetrical trihedral angle are equal.

B'

[merged small][ocr errors]

Let S-A B C be an isosceles trihedral angle, having

LASB=LBSC. And let S-A' B' C be its sym-
metrical trihedral angle.
We are to prove trihedral Z S-ABC=trihedral Z S-A'B'C'.

Revolve < S-A' B' C' about S until S B' falls on S B and the plane S B' A' falls on the plane SBC. Now the dihedral Z C-S B-A = dihedral Z A-S B'-C",

(being vertical dihedral 6).
... the plane S B C will fall on the plane S BA.
Now
ZBSC= L BSA,

Hyp. and

ZB'S A = L BSA,

(being vertical ).
..ZBSC=2 B'SA';

Ax. 1
.. SA will fall on SC.
In like manner S C will fall on SA,
:. the two trihedral { will coincide and be equal.

Q.E.D.

PROPOSITION XXVI. THEOREM. 490. Two symmetrical trihedral angles are equivalent.

Let the trihedral Z S-A BC and Z S-A' B'Cbe sym

metrical.
We are to prove trihedral Z S-ABC =* trihedral Z S-A'B'C'.
Draw D' D making the & DSA, DSC, and D S B equal.
Then L D'SA = _DSC= _ D'SB',

(being vertical És of the equal & D S A, D SC, and DSB).
Then the trihedral Z S-DC B=trihedral Z S-D'C' B', $ 489

(two isosceles symmetrical trihedral e are cqual). And trihedral Z S-DC A = trihedral Z S-D'C' A', . and trihedral Z S-A D B = trihedral Z S-A'D' B'.

Adding the first two equalities, the polyhedral Z S-ABC D = polyhedral Z S-A' B'C'D'.

Take away from each of these equals the equal trihedral E S-A D B and S-A' D' B'. Then trihedral 2 S-A B C = trihedral Z S-A' B'C'.

Q. E. D. 491. Scholium. If D D'fall within the given trihedral angles these trihedral angles would be composed of three isosceles trihedral angles which would be respectively equal, and hence the given trihedral angles would be equivalent.

* The symbol (+) is to be read “ equivalent to."

EXERCISES.

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

2. If each of the projections of a line A B upon two intersecting planes be a straight line, the line A B is a straight line.

3. The height of a room is eight feet, how can a point in the floor directly under a certain point in the ceiling be determined with a ten-foot pole ?

4. If a line be drawn at an inclination of 45° to a plane, what is the greatest angle which any line of the plane, drawn through the point in which the inclined line pierces the plane, makes with the line.

5. Through a given point pass a plane parallel to a given plane.

6. Find the locus of points in space which are equally distant from two given points.

7. Show that the three planes embracing the edges of a trihedral angle and the bisectors of the opposite face angles respectively intersect in the same straight line.

8. Find the locus of the points which are equally distant from the three edges of a trihedral angle.

9. Cut a given quadrahedral angle by a plane so that the section shall be a parallelogram.

10. Determine a point in a given plane such that the sum of its distances from two given points on the same side of the plane shall be a minimum.

11. Determine a point in a given plane such that the difference of its distances from two given points on opposite sides of a plane shall be a maximum.

PROPOSITION XXVII. THEOREM. 492. Two trihedral angles are equal or symmetrical when the three face angles of the one are respectively equal to the three face angles of the other.

[merged small][merged small][merged small][ocr errors][ocr errors]

In the trihedral s S and s', let Z A SB = L A' S' B',

ZASC = 2 A' S' C', and Z BSC = _ B' S' C'.

We are to prove that the homologous dihedral angles are equal, and hence the trihedral angles S and S' are either equal or symmetrical.

On the edges of these angles take the six equal distances SA, S B, SC, S'A', S' B', S'C'.

Draw A B, BC, A C, A'B', B'C', A'C'.

The homologous isosceles A SA B, S'A' B', SAC, S' A' C', SBC, S' B'C' are equal, respectively.

§ 106 .. A B, A C, B C equal respectively A' B', A' C', B'C',

(being homologous sides of equal 6).
..A ABC = A A' B'C'.

§ 108 At any point D in S A draw D E and DFI to SA in the faces A S B and ASC respectively.

These lines meet A B and A C respectively, (since the & S A B and S A C are acute, cach being one of the equal é of an

isosceles A).

Join E F.
On S' A' take A' D' = A D.

Draw D' E' and D' Fi in the faces A' S' B' and A' S' C' respectively I to S' A', and join E' F'. In the rt. A A D E and A' D' E AD= A'D',

Cons. Z DA E= L D'A' E', (being homologous & of the equal A S A B and S' A' B'). .. rt. A ADE =rt. A A' D' E',

§ 111 .. A E = A' E' and D E= D' E',

(being homologous sides of equal A). In like manner we may prove AF = A'F' and DF=D'F'. Hence in the A A E F and A' E' F' we have

A E and A F= respectively A' E' and A' F,
and

ZEAF= 2 E' A' F',
(being homologous of the equal A A BC and A' B'C').
: : AA EF=A A' E' F',

$ 106
..EF= E' F'
(being homologous sides of the equal D A E F and A' E' F').
Hence, in the A EDF and E' D' F' we have
E D, D F, and E F= respectively E' D', D' F', and E' F'.
.: A EDF=A E' D'F',

§ 108 ..EDF=E' D'F',

(being homologous É of equal o ). .: the dihedral 2 B-A S-C = dihedral 2 B'-A'S-C", (since & E D F and E' D' F', the measures of these dihedral &, are equal).

In like manner it may be proved that the dihedral & A-B S-C and A-C S-B are equal respectively to the dihedral A-B'S-C" and A'-CS-B'.

Q. E. D. This demonstration applies to either of the two figures denoted by S'-A' B'C', which are symmetrical with respect to each other. If the first of these figures be given, S and Š are equal, for they can be applied to each other so as to coincide in all their parts. If the second be given, S and S' are symmetrical. § 486

« ΠροηγούμενηΣυνέχεια »