If two straight lines meeting one another be parallel to two other straight lines which meet one another, but are not in the same plane with the first two; the plane which passes through these is parallel to the plane passing through the others.

Let AB, BC (fig. 30), two straight lines meeting one another, be parallel to DE, EF that meet one another, but are not in the same plane with AB, BC: the plane passing through AB, BC shall be parallel to that through DE, EF.

From the point B draw BG perpendicular to the plane which passes through DE, EF (Prop. 5), and let it meet that plane in G; and through G draw GH parallel to ED, and GK parallel to EF (I. 31).

And because BG is perpendicular to the plane through DE, EF, it makes right angles with every straight line meeting it in that plane (Def. 3) but the straight lines GH, GK in that plane meet it; therefore each of the angles BGH, BGK is a right angle: and because BA is parallel to GH (Prop. 19) (for each of them is parallel to DE, and they are not both in the same plane with it), the angles GBA, BGH are together equal to two right angles (Í. 29): and BGH is a right angle; therefore also GBA is a right angle, and GB is perpendicular to BA. For the same reason, GB is perpendicular to BC. Since therefore the straight line GB stands at right angles to the two straight lines BA, BC that cut one another in B, GB is perpendicular to the plane through BA, BC (Prop. 4): and, by construction, it is perpendicular to the plane through DE, EF; therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: but planes to which the same straight line is perpendicular, are parallel to one another (Prop. 25); therefore the plane through AB, BC is parallel to the plane through DE, EF: which was to be proved.

Cor. Through two straight lines AB, EF, not in the same plane, two planes parallel to each other may always be drawn. For taking any point B in AB, and any point E in EF, draw BC parallel to EF, and ED parallel to AB, the plane passing through AB, BC, and that through DE, EF will be parallel.

PROP. XXVII. THEOR.

If two parallel planes be cut by another plane, their common sections with it are parallels.

Let the parallel planes AB, CD (fig. 31) be cut by the plane EFHG, and let their common sections with it be EF, GH EF shall be parallel to GH.

:

For if it be not, EF, GH will meet, if produced, either on the side

of FH, or EG. First, let them be produced on the side of FH, and meet in the point K: therefore, since EFK is in the plane AB, every point in EFK is in that plane (Prop. 1): and K is a point in EFK; therefore K is in the plane AB: for the same reason K is also in the plane CD: wherefore the planes AB, CD produced meet one another: but they do not meet, since they are parallel by the hypothesis; therefore the straight lines EF, GH do not meet when produced on the side of FH. In the same manner it may be proved that EF, GH do not meet when produced on the side of EG. But straight lines which are in the same plane, and do not meet, though produced either way, are parallel; therefore EF is parallel to GH: which was to be proved.

PROP. XXVIII. THEOR.

A straight line which is perpendicular to one of two parallel planes is perpendicular to the other.

Let the straight line AB (fig. 32) be perpendicular to MN, one of the two parallel planes MN, PQ; AB is also perpendicular to the other plane PQ.

From the point B, where the straight line AB meets the plane PQ, draw any straight line BC; then the intersection AD of the plane passing through AB and BC will be parallel to BC (Prop. 26). And since DAB is a right angle (Def. 3), ABC is also a right angle (I. 28). In the same manner it may be shown that AB is at right angles to any other straight line meeting it in the plane PQ: it is therefore (Def. 3) perpendicular to the plane PQ: which was to be proved.

PROP. XXIX. THEOR.

Planes which are parallel to the same plane are parallel to each other. Let each of the planes P, Q (fig. 33) be parallel to the plane R; the planes P and Q are parallel to each other.

From any point C, in the plane R, draw the straight line CBA perpendicular to R (Prop. 7). Then CBA is perpendicular to each of the planes P and Q (Prop. 28); and, consequently, the planes P and Q are parallel (Prop. 25): which was to be proved.

PROP. XXX. THEOR.

Parallel straight lines contained between parallel planes are equal. Let AB, CD (fig. 34) be parallel straight lines contained between the parallel planes MN, PQ: AB and CD are equal.

For AB and CD being parallel, they are in the same plane (I. Def. 35); and the intersections AC, BD of this plane with the parallel planes MN, PQ are parallel (Prop. 27): consequently ABDC

VOL. II.

C

is a parallelogram, and therefore AB and CD are equal: which was to be proved.

PROP. XXXI. THEOR.

If two straight lines be cut by parallel planes, they shall be cut in the

same ratio.

Let the straight lines AB, CD (fig. 35) be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D: as AE is to EB, so shall CF be to FD.

Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel (Prop. 30): for the same reason, because the two parallel planes GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel. And because EX is parallel to BD, a side of the triangle ABD; as AE to EB, so is AX to XD (VI. 2): again, because XF is parallel to AC, a side of the triangle ADC; as AX to XD, so is CF to FD: and it was proved that AX is to XD, as AE to EB; therefore, as AE to EB, so is CF to FD (v. 9). Wherefore, if two straight lines, &c. : which was to be proved.

PROP. XXXII. THEOR.

If a straight line be at right angles to a plane, every plane which passes through it shall be at right angles to that plane.

Let the straight line AB (fig. 36) be at right angles to the plane MN: every plane which passes through AB shall be at right angles to the plane MN.

Let any plane DE pass through AB, and let CE be the common section of the planes DE, MN; take any point F in CE, from which draw FG in the plane DE at right angles to CE (I. 11).

Because AB is perpendicular to the plane MN, therefore it is also perpendicular to every straight line in that plane, meeting it (Def. 3): and consequently it is perpendicular to ĈE: wherefore ABF is a right angle: but GFB by construction is likewise a right angle; therefore AB is parallel to FG (I.28) : and AB is at right angles to the plane MN; therefore FG is also at right angles to the same plane (Prop. 18). But one plane is at right angles to another plane when the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles (Def. 6) to the other plane; and any straight line FG in the plane DE, which is at right angles to CE, the common section of the planes, has been proved to be perpendicular to the other plane MN; therefore the plane DE is at right angles to the plane MN. In like manner, it may be proved that all planes which pass through AB are at right angles to the plane MN: which was to be proved.

Cor. 1. The plane CH being perpendicular to the plane MN, the latter will be perpendicular to the former.

For AB being perpendicular to CE (Def. 3), in the plane MN draw BK at right angles to CE: then AB is at right angles to BK; therefore KB is at right angles to BA and BC, and consequently it is perpendicular to the plane ABC (Prop. 4); and being perpendicular to the common section CE, the plane MN is perpendicular to the plane CH (Def. 6).

Cor. 2. If at a point, three straight lines are severally perpendicular, each to the other two, then each being perpendicular to the plane of the other two, the three planes are perpendicular to each other.

PROP. XXXIII. THEOR.

When two planes are perpendicular to each other, every perpendicular to one of the planes drawn from a point in the other is contained wholly in the latter.

Let the planes MN, CH (fig. 37) be perpendicular to each other, a straight line drawn perpendicular to the plane MN from any point A, in the plane CH will be contained wholly in the plane CH.

For, from A, draw AB perpendicular to CE, the intersection of the two planes: then AB is perpendicular to the plane MN (Def. 6); and besides AB, no other straight line can be drawn from the point A perpendicular to the plane MN (Prop. 6); and AB is wholly in the plane CH: which was to be demonstrated.

Cor. Through a straight line not perpendicular to a plane only one plane can be drawn perpendicular to that plane. For the perpendicular plane must contain the given straight line, and likewise a straight line drawn from a point in it perpendicular to the given plane: and two straight lines which intersect, determine a plane (Prop. 2).

PROP. XXXIV. THEOR.

If two planes which cut one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane.

Let two planes AB, BC (fig. 38) be each of them perpendicular to a third plane MN, and let BD be the common section of the first two: BD shall be perpendicular to the plane MN.

If it be not, from the point D draw, in the plane AB, the straight line DE at right angles to AD (I. 11) the common section of the planes AB and MN; and in the plane BC draw DF at right angles to CD the common section of the planes BC and MN.

And because the plane AB is perpendicular to the plane MN, and DE is drawn in the plane AB at right angles to AD, their com

mon section, DE is perpendicular to the plane MN (Def. 6). In the same manner, it may be proved that DF is perpendicular to the plane MN. Wherefore, from the point D two straight lines stand at right angles to the third plane, upon the same side of it, which is impossible (Prop. 8): therefore, from the point D there cannot be any straight line at right angles to the plane MN, except BD the common section of the planes AB, BC: therefore BD is perpendicular to the plane MN: which was to be proved.

Cor. 1. When three planes are perpendicular to each other, the intersection of any two of these planes is perpendicular to the third plane; and the three intersections are perpendicular to each other.

Cor. 2. A plane perpendicular to two given planes which cut one another is perpendicular to their intersection. For each of the given planes is perpendicular to the third plane; consequently their intersection is perpendicular to that plane, and reciprocally, the third plane is perpendicular to their intersection.

PROP. XXXV. THEOR.

Two planes which are perpendicular to a third plane, and which pass through two parallel straight lines not perpendicular to that plane, are parallel.

Let the two planes AQ, CS (fig. 39) be perpendicular to the plane MN, and pass through the parallel straight lines AB, CD which are not perpendicular to the plane MN: the planes AQ and CS are parallel.

Let PQ and RS be the sections of the planes AQ, CS with the plane MN. In the planes AQ, CS, draw AP, CR perpendicular to PQ, RS; then AP and CR are perpendicular to the plane MN (Def. 6), and are therefore parallel (Prop. 17). And because the straight lines BA, AP, meeting one another, are parallel to DC, CR which meet one another, the planes BAP and DCR are parallel (Prop. 26): which was to be proved.

PROP. XXXVI. THEOR.

Two straight lines, not in the same plane, being given, a straight line which is perpendicular to them both may always be drawn; only one such line can be drawn; and this line is the shortest distance between the two straight lines.

Let AB, CD (fig. 40) be any two straight lines not in the same plane; a straight line which is perpendicular to both AB and CD may be drawn.

Through any point A, in the straight line AB, draw AE parallel to CD (I. 31), and let the plane MN pass through AB and AE.