Then,

But,

which is absurd; hence,

DA = EH = FG Prop.
DA EB = FC

.. EH EB, FG FC

Cor. 1. If two parallel planes XZ, WY, are met by two other planes ADEB, CFEB, the angles ABC, DEF, formed by the intersection of the parallel planes, will be equal.

For the section AB is parallel to the section DE, Prop.

So also, the section BC is parallel to the section EF.

.. angle ABC = angle DEF.

Cor. 2. If three straight lines AD, BE, CF, not situated in the same plane, be equal and parallel, the triangles ABC, DEF, formed by joining the extremities of these straight lines, will be equal, and their planes will be parallel.

PROP. XIV.

If two straight lines be cut by parallel planes, they will be cut in the same

ratio.

Let the straight lines AB, CD, be cut by the parallel planes XZ, WY, VS, in the points A, E, B; C, F, D;

Then, AE: EB:: CF: FD. Join A, C; B, D; A, D; and let AD meet the plane WY in G; join E, G; G, F;

Then, the intersections EG, BD, of the parallel planes WY, VS, with the plane ED, are parallel.

(Prop. x.)

.. AE EB:: AG: GD

Again, the intersections AC, GF, of the parallel planes XZ, YW, with the plane CG, are parallel,

.. AG: GD :: CF: FD

.. comparing this with the first proportion,

AE EB:: CF: FD

PROP. XV.

If a straight line be at right angles to a plane, every plane which passes through it will be at right angles to that plane.

Let the straight line PQ be at right angles to

the plane XZ.

Through PQ draw any plane PO, intersecting XZ in the line OQW.

Then, the plane PO is perpendicular to the plane XZ.

Draw RS, in the plane XZ, perpendicular to WQO.

W

S

P

T

X

Then, since the straight line PQ is perpendicular to the plane XZ, it is perpendicular to the two straight lines RS, OW, which pass through its foot in that plane.

But the angle PQR, contained between PQ, QR which are perpendiculars to OW, the common intersection of the planes XZ, PO, measures the angle of the two planes (Def. 5); hence, since this angle is a right angle, the two planes are perpendicular to each other.

Cor. If three straight lines, such as PQ, RS, OW, be perpendicular to each other, each will be perpendicular to the plane of the two others, and the three planes will be perpendicular to each other.

PROP. XVI.

If two planes be perpendicular to each other, a straight line drawn in one of the planes perpendicular to their common section, will be perpendicular to the other plane.

Let the plane VO be perpendicular to the plane XZ and let OW be their common section.

In the plane VO draw PQ perpendicular to OW;

Then PQ is perpendicular to the plane XZ. From the point Q, draw QR in the plane XZ, perpendicular to OW

Then, since the two planes are perpendicular,

the angle PQR is a right angle.

.. The straight line PQ, is perpendicular to

P

T

W

Q

R

thre straight lines QR, QO, which intersect at its foot in the plane XZ. .. PQ is perpendicular to the plane XZ.

X

Cor. If the plane VO be perpendicular to the plane XZ, and if from any point in OW, their common intersection, we erect a perpendicular to the plane XZ, that straight line will lie in the plane VO.

For if not, then we may draw from the same point a straight line in the plane VO, perpendicular to OW, and this line, by the Prop. will be perpendicular to the plane XZ.

Thus we should have two straight lines drawn from the same point in the plane XZ, each of them perpendicular to the given plane, which is absurd.

PROP. XVII.

If two planes which cut each other, be each of them perpendicular to a third plane, their common section will be perpendicular to the same plane.

Let the two planes VO, TW, whose common section is PQ, be both perpendicular to the plane XZ.

Then, PQ is perpendicular to the plane XZ. For, from the point Q, erect a perpendicular to the plane XZ.

Then, by Cor. to last Prop., this straight line must be situated at once in the planes VO and TW, and is.. their common section.

T

V

P

X

W

2

A solid angle is the angular space contained between several planes which meet in the same point.

Three planes, at least, are required to form a solid angle.

A solid angle is called a trihedral, tetrahedral, &c. angle, according as it is formed by three, four, . . . . plane angles.

PROP. I.

If a solid angle be contained by three plane angles, the sum of any two of these angles will be greater than the third.

It is unnecessary to demonstrate this proposition except in the case where the plane angle, which is compared with the two others, is greater than either

of them.

Let A be a solid angle, contained by the three plane angles BAC, CAD, DAB, and let BAC be the greatest of these angles;

Then, CAD + DAB → BAC.

In the plane BAC draw the straight line AE, making the angle BAE = angle BAD.

D

Make AE AD, and through E draw any straight line BEC, cutting AB, AC, in the points

CE

B, C; join D, B; D, C;

B

Then, AD = AE, and AB is common to the two triangles DAB, BAE, and the angle DAB = angle BAE.

Again, ·· AD = AE, and AC is common to the two triangles DAC, EAC, but the base DC base EC.

..

angle DAC

angle EAC

But, angle DAB = angle BAE

.: angle CAD + angle DAB > angle BAE + angle EAC

> angle BAC.

PROP. II.

The sum of the plane angles which form a solid angle, is always less than four

right angles.

Let P be a solid angle contained by any number of plane angles APB, BPC, CPD, DPE, EPA. Let the solid angle P be cut by any plane ABCDE.

Take any point O in this plane; join A, O; B, 0; C, 0; D, 0; E, 0;

Then, since the sum of all the angles of every triangle is always equal to two right angles, the

E

P

sum of all the angles of the triangles ABP, BPC, ..... about the point P, will be equal to the sum of all the angles of the equal number of triangles about the point 0.

АОВ, ВОС,

Again, by the last Prop., angle ABC like manner, angle BCD

angles of the polygon ABCDE.

angle ABP + angle CBP; in angle BCP + angle DCP, and so for all the

Hence, the sum of the angles at the bases of the triangles whose vertex is 0, is less than the sum of the angles at the bases of the triangles whose vertex is P.

.. The sum of the angles about the point O, must be greater than the sum of the angles about the point P.

But, the sum of the angles about the point O, is four right angles.

.. The sum of the angles about the point P, is less than four right angles.

PROP. III.

If two solid angles be formed by three plane angles which are equal, each to each the planes in which these angles lie will be equally inclined to each other. Let P, Q, be two solid angles, each contained by three plane angles;

Let angle APC = angle DQF, angle APB = angle DQE, and angle BPC = angle EQF.

Then, the inclination of the planes APC, APB, will be equal to the inclination of the planes DQF, DQE

Take any point B in the intersection of the planes APB, CPB.

P

B

D

A

Ꮓ

F

From B draw BY perpendicular to the plane APC, meeting the plane in Y. From Y draw YA, YC, perpendiculars on PA, PC; join A, B; B, C;

Again, take QE = PB, from E draw EZ perpendicular to the plane DQF, meeting the plane in Z, from Z draw ZD, ZF, perpendiculars on QD, QF; join D, E; E, F.

The triangle PAB is right angled at A, and the triangle QDE is right angled at D. (Geom. of Planes, Prop. vII.)

Also, the angle APB = angle DQE, by construction.

.. angle PBA = angle QED

But, the side PB = side QE, .. the two triangles APB, DQF, are equal and similar.

.. PA = QD, and, AB = DE

In like manner, we can prove that,

PC =

QF, and, BC = EF

We can now prove that the quadrilateral PAYC, is equal to the quadrilateral QDZF.

For, let the angle APC be placed upon the equal angle DQF, then the point A will fall upon the point D, and the point C on the point F, because PA=QD, and PC QF.

=

At the same time, AY, which is perpendicular to PA, will fall upon DZ, which is perpendicular to QD; and in like manner, CY will fall upon FZ. Hence, the point Y will fall on the point Z, and we shall have,

AY DZ, and, CY = FZ

But, the triangles AYB, DZE, are right angled in Y and Z, the hypothenuse AB= hypothenuse DE, and the side AY side DZ; hence, these two triangles are equal.

.. angle YAB = angle ZDE

The angle YAB is the inclination of the planes APC, APB; and,
The angle ZDE is the inclination of the planes DQF, DQE.

.. These planes are equally inclined to each other.

In the same manner, we prove that angle YCB = angle ZFE, and consequently, the inclination of the planes APC, BPC, is equal to the inclination of the planes DQF, EQF.

We must, however, observe, that the angle A of the right angled triangle YAB, is not, properly speaking, the inclination of the two planes APC, APB, except when the perpendicular BY falls upon the same side of PA as PC does; if it fall upon the other side, then the angle between the two planes will be obtuse, and, added to the angle A of the triangle YAB, will make up two right angles. But, in this case, the angle between the two planes DQF, DQE, will also be obtuse, and, added to the angle D of the triangle ZDE, will make up two right angles.

Since, then, the angle A will always be equal to the angle D, we infer that the inclination of the two planes APC, APB, will always be equal to the inclination of the two planes DQF, DQE. In the first case, the inclination of the plane is the angle A or D; in the second case, it is the supplement of those angles.

SCHOLIUM. If two solid trihedral angles have the three plane angles of the one equal to the three plane angles of the other, each to each, and at the same time the corresponding angles arranged in the same manner in the two solid angles, then these two solid angles will be equal; and if placed one upon the other, they will coincide. In fact, we have already seen, that the quadrilateral PAYC will coincide with the quadrilateral QDZF. Thus, the point Y falls upon the point Z, and, in consequence of the equality of the triangles AYB, DZE, the straight line YB, perpendicular to the plane APC, is equal to the straight line, ZE perpendicular to the plane DQE; moreover, these perpendiculars lie in the same direction; hence, the point B will fall upon the point E, the straight line PB on the straight line QE, and the two solid angles will entirely coincide with each other.

This coincidence, however, cannot take place, except we suppose the equal plane angles to be arranged in the same manner in the two solid angles; for if the equal plane angles be arranged in an inverse order, or, which comes to the same thing, if the perpendiculars YB, ZE, instead of being situated both on the same side of the planes APC, DQF, were situated on opposite sides of these planes, then it would be impossible to make the two solid angles coincide with each other. It would not, however, be less true, according to the above theorem, that the planes, in which the equal angles lie, would be equally inclined to each other; so that the two solid angles would be equal in all their constituent parts, without admitting of superposition. This species of equality, which is not absolute, or equality of coincidence, has received from Legendre a particular description. He terms it equality of symmetry.

Thus, the two solid trihedral angles in question, which have the three plane angles of the one, equal to the three plane angles of the other, each to each, but