being parallel to BC, is also (XI. 8.) at right angles to the plane through ED, DA; and is therefore perpendicular (XI. def. 1.) to every straight line meeting it in that plane: GH is consequently perpendicular to AF. Therefore AF is perpendi cular to each of the straight lines GH, DE; and consequently (XI. 4.) to the plane BH: wherefore AF is the perpendicular required.

PROP. XII. PROB.

To draw a straight line perpendicular to a given plane, from a point given in the plane.

Let A be the point given in the plane; it is required to draw a perpendicular from A to the plane.

D

B

A

C

From any point B, above the plane, draw (XI. 11.) BC perpendicular to it: if this pass through A, it is the perpendicular required. If not, from A draw (I. 31.) AD parallel to BC. Then, because AD, CB are parallel, and one of them BC is at right angles to the given plane, the other AD is also (XI. 8.) at right angles to it.

1

PROP. XIII. THEOR.

FROM the same point in a given plane, there cannot be two straight lines drawn perpendicular to the plane, upon the same side of it: and there can be but one perpendicular to á plane from a point above it.

B

C

For, if it be possible, let AB, AC, be at right angles to a given plane from the same point A in the plane, and upon the same side of it; and let a plane pass through BA, AC; DAE, the common section of this with the given plane is (XI. 3.) a straight line passing through A. Then (XÍ. def. 1.) because BA is at right angles to the given plane, BAE is a right angle. For the same reason CAE is a right angle. Therefore BAE is equal to CAE; which (I. ax. 9.) is impossible. Also, from a point above a plane, there can be but one perpendicular to that plane; for, if there could be two, they would be parallel (XI. 6.) to one another, which (I. def. 11.) is absurd. Therefore, from the same point, &c.

PROP. XIV. THEOR.

D

K

PLANES to which the same straight line is perpendicular, are parallel to one another.

Let the straight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to one another.

C

If not, they will meet when produced; let them meet; their common section will be a straight line GH, in which take any point K, and join AK, BK. Then because AB is perpendicular to the plane EF, it is perpendicular (XI. def. 1.) to the straight line BK, which is in that plane. Therefore ABK is a right angle. For the same reason BAK is a right angle; wherefore, in the triangle ABK, the two angles ABK, BAK are equal to two right angles, which (I. 17.) is impossible. Therefore the planes CD, EF, though produced, do not meet one another; that is, (XI. def. 4.) they are parallel. Therefore planes, &c.

PROP. XV. THEOR.

H

B

E

D

Two planes are parallel, if two straight lines which meet one another on one of them, be parallel to two which meet on the other.

Let the straight lines AB, BC meet on the plane AC, and DE, EF, on the plane DF: if AB, BC be parallel to DE, EF, the plane AC is parallel to DF.

B

Ꭺ

E

F

G

K

C

DH

From B draw (XI. 11.) BG perpendicular to the plane DF, and let it meet that plane in G; and through G draw (I. 31.) GH parallel to ED, and GK to EF. Then, because BG`is perpendicular to the plane DF, each of the angles BGH, BGK is (XI. def. 1.) a right angle: and because (XI. 9.) BA is parallel to GH, each of them being parallel to DE, the angles GBA, BGH are together equal (I. 29.) to two right angles. But BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA. For the same reason, GB is perpendicular to BC. Since therefore GB is perpendicular to BA, BC, it is perpendicular (XI. 4.) to the plane AC: and (const.) it is perpendicular to the plane DF. But (XI. 14.) planes to which the same straight line is perpendicular are parallel to one another: therefore the planes AC, DF are parallel. Wherefore two planes, &c.

PROP. XVI. THEOR.

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

Let the parallel planes, AB, CD, be cut by the plane EFHG, and let their common sections with it be EF, GH: EF is parallel to GH.

K

H

D

For, if it be not, EF, GH will meet if produced, either towards FH or EG. First, let them be produced towards FH, and meet in the point K. Then, since EFK is in the plane AB, K is in AB. For the same reason, K is also in CD: wherefore the planes AB, CD produced meet one another; but they do not meet, since (hyp.) they are parallel. Therefore EF, GH do not meet when produced through F, H: and it may be proved in the same manner, that they do not meet when produced through E, G: they are therefore (I. def. 11.) parallel. Wherefore, if two parallel planes, &c.

PROP. XVII. THEOR.

A

G

IF two straight lines be cut by parallel planes, they are 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 ; C, F, D: as AE: EB:: CF: FD.

G

A

C

H

L

P

B

Join AC, BD, AD, and let AD meet KL in X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are (XI. 16.) parallel. For the same reason, because GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel. Then, (VI. 2.) because EX is parallel to BD, a side of the triangle ABD, AE: EB:: AX: XD; and, because XF is parallel to AC, a side of the triangle ADC, AX: XD :: CF: FD: and it was proved that AX: XD:: AE: EB; therefore (V. 11.) AE EB:: CF: FD. Wherefore, if two straight lines, &c.

PROP. XVIII. THEOR.

५

N

D

If a straight line be perpendicular to a plane, every plane which passes through it is perpendicular to that plane.

Let AB be perpendicular to the plane CK; every plane which passes through AB is perpendicular to CK.

D G A

H

Let any plane DE pass through AB, and let CE be its common section with CK; take any point F in CE, from which draw FG in the plane DE perpendicular to CE. Then, because AB is perpendicular to CK, it is perpendicular (XI. def. 1.) to CE, which meets it in that plane. Therefore ABF is a right angle; but (const.) GFB

is likewise a right angle: therefore (1. 28.) AB is parallel to FG.

D G A

H

K

Now AB is at right angles to the plane CK; therefore (XI. 8.) FG is also at right angles to the same plane. But (XI. def. 3.) one plane is at right angles to another, when the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles 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 CK: therefore DE is at right angles to CK. In like manner, it may be proved, that any other plane passing through AB is perpendicular to CK. Therefore, if a straight line, &c.

PROP. XIX. THEOR.

C

F B E

If two planes cutting one another be each perpendi-, cular to a third plane; their common section is perpendicular to the same plane.

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

B

P

If it be not, from D draw, in AB, the straight line DE perpendicular to AD, the common section of AB with ADC; and in BC draw DF perpendicular to CD, the common section of BC with ADC. Then, because AB is perpendicular to ADC, and DE is drawn in AB at right angles to AD their common section, DE (XI. def. 3.) is perpendicular to ADC. In the same manner it may be proved that DF is perpendicular to ADC. Therefore, from the point D two straight lines are drawn at right angles to the third plane, upon the same side of it, which (XI. 13.) is impossible: from D, therefore, there cannot be any perpendicular to the third plane, except BD the common section of the planes AB, BC; BD, therefore, is perpendicular to the third plane. Wherefore, if two planes, &c.

PROP. XX. THEOR.

Ir 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 these are greater than the third. If the angles be all equal, it is evident that any two of them are greater than the third. But if they be not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and make, in the plane of BA, AC, the angle BAE equal (I. 23.) to DAB; make AE equal to AD; through È draw BEC cutting AB, AC, in

B

E C

the points B, C, and join DB, DC. Then, in the triangles BAD, BAE, because DA is equal to AE, AB common, and the angle DAB is equal to EAB; DB is equal (I. 4.) to BE. Again, because (I. 20.) BD, DC are greater than CB, and one of them BD has been proved equal to BE a part of CB, therefore the other DC is greater (I. ax. 5.) than the remaining part EC. Then, because DA is equal (const.) to AE, and AC common, but the base DC greater than the base EC; therefore (I. 25.) the angle DAC is greater than EAC; and (const.) the angles DAB, BAE are equal; wherefore (I. ax. 4.) the angles DAB, DAC are together greater than BAE, EAC, that is, than BAC. But BAC is not less than either of the angles DAB, DAC; therefore BAC, with either of them, is greater than the other. Wherefore, if a solid angle, &c.

PROP. XXI. THEOR.

EVERY solid angle is contained by plane angles, which together are less than four right angles.

Let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; these together are less than four right angles.

B

P

Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB. Then, because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which (XI. 20.) any two are greater than the third, CBA, ABF are greater than FBC. For the same reason, the two plane angles at each of the points C, D, E, F, viz., the angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the angles of the figure BCDEF. Therefore all the angles at the bases of the triangles are together greater than all the angles of that figure: and because (I. 32.) all the angles of the triangles are together equal to twice as many right angles as there are triangles; that is, as there are sides in the figure BCDEF; and that (I. 32. cor. 1.) all the angles of the figure, together with four right angles, are likewise equal to twice as many right angles as there are sides in the figure; therefore all the angles of the triangles are equal to all the angles of the figure, together with four right angles. But all the angles at the bases of the triangles are greater than all the angles of the figure, as has been proved: wherefore the remaining angles of the triangles, viz., those at the vertex, which contain the solid angle at A, are less than four right angles. Therefore every solid angle, &c.

C

D

Schol. This proposition does not necessarily hold, if any of the