.: GHFE is a parallelogram; and therefore,

EG FH

Cor. Two parallel planes are every where equidistant.

PROP. XII.

If two planes be parallel to each other, a straight line which is perpendicular to one of the planes, will be perpendicular to the other also.

Let the two planes XZ, WY, be parallel, and let the straight line AB, be perpendicular to the plane XZ;

Then, AB will be perpendicular to WY.

For, from any point H in the plane WY, draw HG perpendicular to the plane XZ, and draw AG, BH.

Then, since BA, HG, are both perpendicular to XZ; .. the angles A, G, are right angles.

W

H

Y

And, since the planes XZ, WY, are parallel, .. the perpendiculars BA, HG, are equal.

Hence AG is parallel to BH, and AB being perpendicular to AG, is perpendicular to BH also.

In like manner, it may be proved, that AB is perpendicular to all other lines which can be drawn from B in the plane WY.

.. AB is perpendicular to the plane WY.

Cor. Conversely, if two planes be perpendicular to the same straight line, they will be parallel to each other.

PROP. XII.

If two straight lines which form an angle, be parallel to two other straight lines which form an angle in the same direction, although not in the same plane with the former, the two angles will be equal, and their planes will be parallel. Let the two straight lines AB, BC, in the plane XZ, be parallel to the two DE, EF, in the plane WY; Then, angle ABC = angle DEF.

For, make AB = DE, BC= EF; join A, C; D, F; A, D; B, E; C, F;

Then, the straight lines AD, BE, which join the equal and parallel straight lines AB, DE, are themselves equal and parallel.

lel.

For the same reason, CF, BE, are equal and paral

Z

W

C

H

X

.: AD, CF, are equal and parallel, and .. AC, DF, are, also, equal and parallel.

Hence, the two triangles ABC, DEF, having all their sides equal, each to each, have their angles also equal.

.. angle ABC = angle DEF.

Again, the plane XZ is parallel to the plane WY.

For, if not, let a plane drawn through A parallel to DEF, meet the straight lines FC, EB, in G and H.

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

AE EB: CF: FD

.. comparing this with the first proportion,

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.

T

P

V

S

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

Ꮓ

P

T

W

Q

R

the 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

Р

V

X

W

SOLID ANGLES.

DEFINITION.

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.

Make AE AD, and through E draw any straight line BEC, cutting AB, AC, in the points B, C; join D, B; D, C;

CE

D

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.

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

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 AOB, BOC, about the point O.

............

Again, by the last Prop., angle ABC angle ABP + angle CBP; in like manner, angle BCDangle BCP+ angle DCP, and so for all the angles of the polygon ABCDE.

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.

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