are parallel to each other. For, conceive a plane perpendicular to the line C: the lines A and B, being parallel to C, will be perpendicular to the same plane; therefore, by the preceding corollary, they will be parallel to each other.

When the three lines are in the same plane, the case corresponds with Book I, Prop. xxi.

PROPOSITION VIII.

THEOREM. If a straight line without a plane is parallel to a line within the plane, it is parallel to the plane itself.

Let the straight line AB, without the plane MN, be parallel to the line CD of this plane; then will AB be parallel to the plane MN.

N

B

D

For if the line AB, which lies in the plane ABDC, could meet the plane MN, this could only be in some point of the line CD, the intersection of the two planes; but AB cannot meet CD, since they are parallel; hence it will not meet the plane MN; therefore (Def. 5,) it is parallel to that plane.

PROPOSITION IX.

THEOREM. If two planes are perpendicular to the same line, they are parallel.

Let the planes MN and PQ be each perpendicular to AB; then will they be parallel. For, if they can meet anywhere, let Q O be one of their

M

A

N

B

common points, and join OA, OB. The line AB, which is perpendicular to the plane MN, is perpendicular to the straight line OA drawn through its foot in that plane. For the same reason, AB is perpendicular to BO. Therefore OA and OB are two perpendiculars drawn from the same point O, upon the same straight line, which is impossible; hence the planes MN, PQ cannot meet each other, and consequently they are parallel.

PROPOSITION X.

THEOREM. The intersections of two parallel planes with a third plane are parallel.

Let the two parallel planes MN and PQ intersect the plane EH; then will EF be parallel to HG.

For, if the lines EF, GH, lying in the same plane, were not parallel, they would meet

H

M

each other when produced; therefore the planes MN,

PQ, in which those lines are situated, would also meet therefore the planes would not be parallel.

PROPOSITION XI.

THEOREM. A line which is perpendicular to one of two parallel planes, is perpendicular to the other also.

Let the two planes MN and PQ be parallel; then if the line AB is perpendicular to the plane MN, it will also be perpendicular to PQ.

N

B

M

-D

Having drawn any line BC in the plane PQ, along the lines AB and BC, extend a plane ABC, intersecting the plane MN in AD; the intersection AD will be parallel to BC, (B. VI, Prop. x.) But the line AB, being perpendicular to the plane MN, is perpendicular to the straight line AD; therefore also it is perpendicular to its parallel BC, (B. I, Prop. xvii, Cor.) Hence the line AB, being perpendicular to any line BC drawn through its foot in the plane PQ, is consequently perpendicular to that plane.

PROPOSITION XII.

THEOREM. Two parallel lines, included between two parallel planes, are equal.

Let EG, FH be two parallel lines included between the two parallel planes MN, PQ; then will these lines be equal.

Through the parallels EG, FH, draw the plane EGHF to meet the parallel planes in EF

and GH. The intersections EF, GH, (B. VI, Prop. x,) are parallel to each other; so likewise are EG, FH: therefore the figure EGHF is a parallelogram, and EG=FH.

Cor. Hence it follows that two parallel planes are everywhere equidistant; for if EG and FH are perpendicular to the two planes MN, PQ, they will be parallel to each other, (B. VI, Prop. vII, Cor. 1,) and therefore equal.

PROPOSITION XIII.

THEOREM. 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 an

F

B

gles be equal, and their planes will be parallel.

M

A

P

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, Prop. xxix ;) 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 CAE DBF.

=

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 (B. VI, Prop. 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 CABD, EABF, the angles CAE, DBF formed by the intersections of the parallel planes will be equal; for (B. VI, Prop. x,) the intersection AC is parallel to BD, and AE to BF, and therefore the angle CAE=DBF.

PROPOSITION XIV.

THEOREM. 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.