PROPOSITION XIII. THEOREM

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

Let CAE and DBF be two angles lying in the planes MN and PQ, and let the sides AC and AE be respectively parallel to BD and BF, and lying in the same direction then will the angles CAE and DBF be equal, and the planes MN and PQ will be parallel.

Take any two points of AC and AE, as C and E, and make BD equal to AC, and

BF to AE; draw CE, DF,

AB, CD, and EF

1°. The angles CAE

DBF will be equal.

H

G

and

A

E

P

B

For, AE and BF being parallel and equal, the figure ABFE is a parallelogram (B. I., P. XXX.); hence, EF is parallel and equal to AB. For a like reason, CD is parallel and equal to AB: hence, CD and EF are parallel and equal to each other, and consequently, CE and DF are also parallel and equal to each other. The triangles CAE and DBF have, therefore, their corresponding sides equal, and consequently, the corresponding angles CAE and DBF are equal; which was to be proved.

2. The planes of the angles MN and PQ are parallel. For, if not, pass a plane through A parallel to PQ, and suppose it to cut the lines CD and EF in G and H Then will the lines GD and HF be equal respect

ively to AB (P. XII.), and consequently, GD will be equal to CD, and HF to EF; which is impossible: hence, the planes MN and PQ must be parallel; which was to be proved.

Cor. If two parallel planes MN and PQ, are met by two other planes AD and AF, and AF, the angles CAE and DBF, formed by their intersections, will be equal.

PROPOSITION XIV. THEOREM.

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

Let AB, CD, and EF be equal parallel lines not in the same plane: then will the triangles ACE and BDF be equal, and their planes parallel.

For, AB being equal and parallel to EF, the figure ABFE is a parallelogram, and consequently, AE is equal and parallel to BF. For a like reason, AC is equal and parallel to BD: hence, the included angles CAE and DBF are equal and their planes parallel (P. XIII.).

Now, the triangles CAE and
DBF have two sides and their

included angles equal, each to each

D

A

E

hence, they are equal

in all their parts. The triangles are, therefore, equal and

their planes parallel; which was to be proved.

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

Let the lines AB and CD be cut by the parallel lanes MN, PQ, and RS, in the points A, E, B, and C, F, D; then

AE EB :: CF : FD.

For, draw the line AD, and suppose it to pierce the plane PQ in G; draw AC, BD, EG, and GF.

The plane ABD intersects the parallel planes RS and PQ in the lines BD and EG; consequently, these lines are parallel (P. X.): hence (B. IV., P. XV.),

AE : EB :: AG : GD.

M

S

B

R

P

F

The plane ACD intersects the parallel planes MN and PQ, in the parallel lines AC and GF: hence,

AG: GD :: CF: FD.

Combining these proportions (B. II., P. IV.), we have,

AE EB :: CF: FD;

which was to be proved.

Cor. 1. If two straight lines are cut by any number of parallel planes, they will be divided proportionally.

Cor. 2. If any number of straight lines are cut by three parallel planes, they will be divided proportionally.

If a straight line is perpendicular to a plane, every plane passed through the line will also be perpendicular to that plane.

Let AP be perpendicular to the plane MN, and let BF be a plane passed through AP: then will BF be perpendicular to MN.

F

N

is perpendicular to MN, it is perpendicular to BC and DP (D. 1);

DP, in the

M

and since AP and planes BF and MN, are perpendicular to the intersection of these planes at the same point, the angle which they form is equal to the angle formed by the planes (D. 4); but this angle is a right angle: hence, BF is perpendicular to MN; which was to be proved.

Cor. If three lines. AP, BP, and DP, are perpendicular to each other at a common point P, each line will be perpendicular to the plane of the other two, and the three planes will be perpendicular to each other.

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

Let the planes BF and MN be perpendicular to each other, and let the line AP, drawn in the plane BF, be perpendicular to the intersection BC; then will AP be perpendicular to the plane MN.

For, in the plane MN, draw PD perpendicular to BC Then because the planes BF and MN are perpen

at P.

Cor. If the plane BF is perpendicular to the plane MN, and if at a point P of their intersection, we erect a perpendicular to the plane MN, that perpendicular will be in the plane BF. For, if not, draw in the plane BF, PA perpendicular to PC, the common intersection ; AP will be perpendicular to the plane MN, by the theorem ; therefore, at the same point P, there are two perpendiculars to the plane MN; which is impossible (P. IV., C. 2).

PROPOSITION XVIII. THEOREM.

If two planes cut each other, and are perpendicular to a third plane, their intersection is also perpendicular to that plane.

H

F

A

Let the planes BF, DH, be perpendicular to MN: then will their intersection AP be perpendicular to MN. For, at the point P, erect a perpendicular to the plane MN; that perpendicular must be in the plane BF, and also in the plane DH (P. XVII., C.); therefore, it is their common intersection AP; which was to be proved.

B

M

N