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 make BD equal to AC, and BF to AE; draw CE, DF, AB, CD, and EF.

1o. The angles CAE

DBF will be equal.

and AE, as C and E, and

H

G

E

M

P

F

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 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.

2o. 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 Then will the lines GD and HF

H.

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 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

D

F

included angles equal, each to each: 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 planes MN, PQ, and RS, in the points A, E, B,

B, and

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

AG: GD :: CF: FD.

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

AE : EB :: : CF
CF : FD;

which was to be proved.

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

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

If a 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

and since AP and DP,
AP and
planes BF and MN,

in the

DP,

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 dicular to each other at

be perpendicular to the

AP, BP,
BP, and DP, are perpen-

a common point P,
common point P, each line will
plane of the other two, and the

three planes will be perpendicular to each other.

If two planes are perpendicular to each other, a 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 at P. Then because the planes BF and MN are perpendicular to each other, the angle APD

will be a right angle: hence, AP is perpendicular to the two lines PD and BC, at their intersection, and consequently, is perpendicular to their plane MN; which was to be proved.

F

A

C

B P

D

M

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 be in the plane BF PA perpendicular to

plane MN, that perpendicular will For, if not, draw in the plane BF

PC,

will be perpendicular to the

the common intersection; AP 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, DII, 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