THEOREM XIII.

If two angles, not situated in the same plane, have their sides parallel and lying in the same direction, they will be equal, and the planes in which they are situated will be parallel.

Let CAE, DBF be two angles not situated in the same plane, having AC parallel to BD and lying in the same direction, and AE parallel to BF, and also lying in the same direction; then will these angles be equal, and their planes will be parallel..

N

M

C

E

A

P

D

F

B

Make AC = BD, AE = BF; and join CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABDC is a parallelogram (B. I., T. XXX.), therefore CD is equal and parallel to AB. For a similar reason, EF is equal and parallel to AB; hence, also, CD is equal and parallel to EF. The figure CEFD is therefore a parallelogram, and the side CE is equal and parallel to DF; therefore the triangles CAE, DBF have their corresponding sides equal, and consequently the angle CAEDBF.

Again, the plane ACE is parallel to the plane BDF. For, suppose the plane parallel to BDF, drawn through the point A, were to meet the lines CD, EF in points different from C and E, for instance in G and H, then (T. XII.) the three lines AB, GD, HF would be equal. But the lines AB, CD, EF are already known to be equal, hence CD = GD, and HF = EF, which is absurd; hence the plane ACE is parallel to BDF.

Cor. If two parallel planes MN, PQ are met by two other planes not parallel CABD, EABF, the angles CAE, DBF formed by the intersections of the parallel planes will be equal; for (T. X.) the intersection AC is parallel to BD, and AE to BF, and therefore the angle CAEDBF.

THEOREM XIV.

If three straight lines, not situated in the same plane, are equal and parallel, the triangles formed by joining their corresponding opposite extremities will be equal, and their planes will be parallel.

Let ACE, BDF be two triangles formed by joining the opposite extremities of the three equal and parallel lines AB, CD, EF; then will these triangles be equal, and their planes will be parallel.

૨

For, since AB is equal and parallel to CD, the figure ABDC is a parallelogram (B. I., T. XXX.); hence the side AC is equal and parallel to BD. For a like reason, the sides AE, BF are equal and parallel, as also CE, DF. Therefore the two triangles ACE, BDF are equal; and, consequently, as in the last proposition, their planes are parallel.

THEOREM XV.

If two straight lines are cut by three parallel planes, they will be divided proportionally.

Suppose the line AB to be cut by the parallel planes MN, PQ, RS, at the points A, E, B; and the line CD to be cut by the same planes at the points C, F, D; then

AE: EB:: CF: FD.

Draw AD meeting the plane PQ in

S

N

M

C

Α

P

F/G

E

R

B

G, and join AC, EG, GF, BD. The intersections EG, BD, of the parallel planes PQ, RS in the plane ABD, are parallel (T. X.); therefore,

AE: EB:: AG: GD (B. III., T. III.).

In like manner, the intersections AC, GF being parallel,

AG: GD:: CF: FD.

The ratio AG: GD is the same in both; hence

AE: EB:: CF: FD.

THEOREM XVI.

If a straight line is perpendicular to a plane, then every plane passing through this line will also be perpendicular to the first plane.

Let the line AP be perpendicular to the plane MN; then any plane, as AB, passing through this line, will also be perpendicular to the plane MN.

For, let BC be the intersection of the planes AB, MN. In the plane MN draw DE perpendicular to BP; then the line AP, being perpendicular to the plane MN, will be perpendicular to each of the two straight lines BC, DE. But the angle APD, formed by the two perpendiculars PA, PD at their common intersection BP, is the measure of the angle of the two planes (D. VI.); and since, in the present case, the angle is a right angle, the two planes are perpendicular to each other.

Scholium. When the three lines, such as AP, BP, DP, are perpendicular to each other, each of these lines-is perpendicular to the plane of the other two, and the planes themselves are perpendicular to each other.

THEOREM XVII.

If two planes are perpendicular to each other, every line drawn in one of them perpendicular to their common intersection, will be perpendicular to the other plane.

Let the planes AB, MN be perpendicular to each other; and in the plane AB, let PA be drawn perpendicular to the common intersection PB; then will it be perpendicular to the plane MN.

N

A

D

M

B

For, in the plane MN draw PD perpendicular to PB; then because the planes are perpendicular, the angle APD is a right angle; therefore the line AP is perpendicular to the two straight lines PB, PD, and consequently perpendicular to their plane MN.

Cor. If the plane AB be perpendicular to the plane MN, and if at a point P of the common intersection a perpendicular be drawn to the plane MN, that perpendicular will be in the plane AB. For, if not, then in the plane AB we might draw AP perpendicular to PB their common intersection, and this AP at the same time would be perpendicular to the plane MN; therefore, at the same point P there would be two perpendiculars to the plane MN, which is impossible.

THEOREM XVIII.

If two planes be perpendicular to a third plane, their common intersection will be also perpendicular to the third plane.

Let AB, AD be perpendicular to MN; then will their common intersection AP be perpendicular to the same plane MN.

For, at the point P draw the perpendicular

to the plane MN; then that perpendicular N

D

M

B

must be in the plane AD; it must also be in the plane AB (T. XVII.), therefore it is their common intersection AP.

THEOREM XIX.

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

The proposition requires demonstration only when the plane angle, which is compared to the sum of the other two, is greater than either of them. Therefore, suppose the polyedral angle S to be formed by three plane angles ASB, ASC, BSC, whereof the angle ASB is the greatest: we are to show that ASB ASC + BSC.

B

C

D

A

In the plane ASB make the angle BSD = BSC; draw the straight line ADB at pleasure; and, having taken SC=SD, join AC, BC.

The two sides BS, SD are equal to the two sides BS, SC; the angle BSD = BSC; therefore the triangles BSD, BSC are equal, and BD = BC. But AB < AC+ BC; taking BD from the one side, and from the other its equal BC, there remains AD < AC. The two sides AS, SD are equal respectively to the two AS, SC; the third side AD is less than the third side AC; therefore the angle ASD < ASC. Adding BSD = BSC, we shall have ASD +BSD, or ASB < ASC + BSC.

THEOREM XX.

The sum of the plane angles which form any polyedral angle, is always less than four right angles.

Conceive the polyedral angle S to be cut by any plane ABCDE; from O a point in that plane, draw to the several angles straight lines AO, OB, OC, OD, OE.

D

C

S

B

A

The sum of the angles of the triangles ASB, BSC, etc., formed about the vertex S, is equivalent to the sum of the angles of an equal number of triangles AOB, BOC, etc., formed about the point O. But at the point B the angles ABO, OBC, taken together, make the angle ABC (T. XIX.) less than the sum of the angles ABS, SBC. In the same manner, at the point C we have BCO + OCD < BCS + SCD, and so with all the angles of the polygon ABCDE. Whence it follows that the sum of all the angles at the bases of the triangles whose common vertex is in O, is less than the sum of all the angles at the bases of the triangles whose common vertex is in S; hence, to make up the deficiency, the sum of the angles formed about the point O, is greater than the sum of the angles about the point S. But the sum of the angles about the point O is equal to four right angles (B. I., T. I., C. III.); therefore the sum of the plane angles, which form the polyedral angle S, is less than four right angles.

Scholium. This demonstration is founded on the supposition that the polyedral angle is convex, or that the plane of no one surface produced can ever meet the polyedral angle. If it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude.

THEOREM XXI.

If two polyedral angles are composed of three plane angles respectively equal to each other, the planes which contain the equal angles will be equally inclined to each other.

Let the angle ASC = DTF, the

=

angle ASB DTE, and the angle BSC ETF; then will the inclination

of the planes ASC, ASB be equal to that of the planes DTF, DTE.

Having taken SB at pleasure, draw

E

T

S

A

D

P

BO perpendicular to the plane ASC; from the point O at which