PROP. XV. THEOR.

IF two parallel planes be cut by a third plane, they have the same inclination to that plane.

Book II.

Let AB, CD be two parallel planes, and EH a third plane cutting them; the planes AB, CD are equally inclined to EH. Let the straight lines EF, GH be the common section of the plane EH with the two planes AB, CD; from K, any point in EF, draw in the plane EH the straight line KM at right angles to EF, and let it meet GH in L; draw also KN at right angles to EF in the plane AB; through the straight lines KM, KN let a plane be made to pass, cutting the plane CD in the line LO. Because EF, GH are the common sections of the plane EH with the two parallel planes AB, CD, EF is parallel to GHa. But EF is at right angles to the a 14.2.Sup. plane that passes through KN and KMb; therefore GH is also at right angles to the same planec, and therefore is at right angles to the lines LM, LO, which it meets in that

plane. Therefore,

since LM and

B

E

Ꮐ

b 4. 2. Sup.

A

C

L

c 7. 2. Sup.

K

M

IN D

0

F

H

LO are at right angles to LG, the common section of the two planes CD and EH, the angle OLM is the inclination of the plane CD to the plane EH. For the same reason the d Def. 4. angle MKN is the inclination of the plane AB to the plane 2. Sup. EH. But KN and LO, being the common sections of the parallel planes AB and CD with a third plane, are parallela; therefore the interior angle NKM is equal to the exterior angle OLM; that is, the inclination of the plane AB to the e 29. 1.

Suppl. plane EH is equal to the inclination of the plane CD to the same plane EH. Therefore, if two parallel planes, &c. Q. E. D.

PROP. XVI. THEOR.

IF two straight lines be cut by parallel planes they will be cut in the same ratio.

Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B, and C, F, D;

PROP. XVII. THEOR.

IF a straight line be at right angles to a plane, every plane which passes through the line is also at right angles to that plane.

Let the straight line AB be at right angles to a plane CK; every plane which passes through AB is at right angles to the plane CK.

Book II.

D

A H

a Def. 1. 2. Sup.

K

Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; take any point F in CE, from which draw FG in the plane DE at right angles to CE. Because AB is perpendicular to the plane CK, it is also perpendicular to every straight line meeting it in that planea; consequently it is perpendicular to CE; wherefore ABF is a right angle. But GFB is a right angle; therefore AB is parallel to FG. But AB is at right angles to the plane CK; therefore FG is also at right angles to CK. Now 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 CK; therefore the plane DE is at right angles to the plane CK. In like manner it may be proved that all the planes which pass through AB are at right angles to the plane CK. Therefore, if a straight line, &c. Q. E. D.

PROP. XVIII. THEOR.

F

B E b 28. 1.

IF two planes which cut each other be perpendicu lar to a third plane, their common section will be perpendicular to the same plane.

Suppl.

a Def. 2. 2. Sup.

b Def. 1.

2. Sup.

Let the two planes AB, BC be perpendicular to a third plane ADC, and let BD be the common section of AB, BC; BD is perpendicular to the plane ADC.

B

From D in the plane ADC draw DE perpendicular to AD, and DF to DC. Because DE is perpendicular to AD, the common section of the planes AB and ADC, and because the plane AB is at right angles to ADC, DE is at right angles to the plane AB2, and therefore also to the straight line BD in that planeb. For the same reason DF is at right angles to DB. Since BD is at right angles to both the lines DE and DF, it is at right angles to the plane in which DE and

c 4.2. Sup. DF are, that is, to the plane ADC. Wherefore, if two planes, &c. AF Q. E. D.

E

PROP. XIX. PRÓB.

TWO straight lines not in the same plane being given in position, to draw a straight line perpendicular to both.

cular to the plane which passes through EB, EF. Through Book II. AB and EG let a plane GK pass, meeting CD in H; and from H draw HK perpendicular to AB; HK is the line required.

Through H draw HG parallel to AB. Since HK, GE, which are in the same plane, are both at right angles to the straight line AB, they are parallel to each other. And be- a 28. 1. cause the lines HG, HD are parallel to the lines EB, EF, each to each, the plane GHD is parallel to the plane BEF; b 9.2. Sup. therefore EG, which is perpendicular to the plane BEF, is perpendicular also to the plane GHD. Therefore HK, which c Cor. 13. is parallel to GE, is also perpendiculard to the plane GHD, 2. Sup. and is therefore perpendiculare to HD, which is in that plane. d 7.2.Sup. But HK is perpendicular to AB; therefore HK is drawn e Def. 1. 2. Sup. perpendicular to the two given lines AB, CD. Which was to be done.

PROP. XX. THEOR.

IF a solid angle be contained by three plane angles, any two of them are greater than the third.

Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB; any two of them are greater than the third.

D

If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not equal, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; at the point A, in the straight line AB, make in the plane which passes through BA, AC the angle BAE equal to the angle DAB; make AE equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. Because DA is equal to AE, and AB is common to the two triangles ABD, ABE, and also the angle DAB

B

a 23. 1.

E C