is at right angles to the plane CK, therefore it makes right angles with every straight line which meets it in that plane; therefore it makes right angles with CE, that is, ABF is a right angle: But GFB is also a right angle; therefore FG is parallel to AB: And AB is at right angles to the plane CK; therefore (11. 8) FG is also at right angles to the same plane.

But (11. Def. 4) one plane is at right angles to another plane, when any straight line drawn in one of the planes, at right angles to their common section, is also at right angles to the other plane; and it has been shewn that any straight line FG drawn in the plane DE, at right angles to CE, the common section of the two planes, is also at right angles to the plane CK— therefore the plane DE is at right angles to the plane CK.

And, in like manner, it may be shewn that any other plane passing through the straight line AB is at right angles to the plane CK.

Wherefore, If a straight line &c.

Q. E. D.

PROP. XIX. THEOR.

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

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

For, if not, from the point D draw, in the plane AB, the straight line DE at right angles to AD, the common section of the plane AB with the third plane; and from the same point draw, in the plane BC, the straight line

B

DF at right angles to CD, the common section of the plane BC with the third plane: Then, because the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD their common section, DE is perpendicular to the third plane (11. Def. 4): And, in like manner, it may be proved that DF is perpendicular to the third plane: Therefore, from the point D, two straight lines stand at right angles to the A

third plane, upon the same side of it—which (11. 13) is impossible: Therefore, from the point D, there cannot be any straight line at right angles to the third plane, except BD the common section of the planes AB, BC: Therefore BD is perpendicular to the third plane. Wherefore, If two planes &c. Q. E. D.

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

Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB: any two of these shall be greater than the third.

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

B

F

Then, because AD is equal to AE, and AB is common, the two AB, AD, are equal to the two AB, AE, each to each, and the angle BAD is equal to the angle BAE-therefore the base BD is equal to the base BE: And because BD, DC are together greater than BC, and one of them BD has been proved equal to BE a part of BC, therefore the other DC is greater than the remaining part EC: Then, because AD is equal to AE, and AC common, but the base DC greater than the base EC, therefore (1. 25) the angle DAC is greater than the angle EAC: And, by the construction, the angle BAD is equal to the angle BAE; therefore the angles BAD, DAC are together greater than the angles BAE, EAC, that is, than the angle BAC: But BAC is not less than either of the angles BAD, DAC; therefore BAC, with either of them, is greater than the other.

Wherefore, If a solid angle &c.

PROP. XXI.

Q. E.D.

THEOR.

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

First, let the solid angle at A be contained by three plane angles BAC, CAD, DAB: these shall be together less than four right angles.

In the straight lines AB, AC, AD, take any points B, C, D, and join BC, CD, DB: Then, because the solid angle at B is contained by the three plane angles CBA, ABD, DBC, any two of these are together greater than the third; therefore the angles CBA, ABD are together greater than the angle CBD : For the same reason, the angles BDA, ADC are together greater than the angle BDC, and the angles DCA, ACB are

D

together greater than the angle DCB: Therefore the six angles CBA, ABD, BDA, ADC, DCA, ACB are together greater than the three angles CBD, BDC, DCB: But the three angles CBD, BDC, DCB are together equal to two right angles; therefore the six angles CBA, ABD, BDA, ADC, DCA, ACB are together greater than two right angles : And, because the three angles of each of the triangles ABC, ACD, ADB are together equal to two right angles, therefore the nine angles of these three triangles are together equal to six right angles : And of these, the six angles CBA, ABD, BDA, B ADC, DCA, ACB are together greater than two right angles; therefore the remaining three angles BAC, BAD, CAD, which contain the solid angle at A, are together less than four right angles.

Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB: these shall be together less than four right angles.

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, any two of which are greater than the third, B therefore the angles CBA, ABF are together greater than the angle FBC: In like manner, at each of the points C, D, E, F, the two angles, which are at the bases of the triangles having the common vertex A, are together greater than the third angle at the same point, which is one of the angles of the polygon BCDEF: Therefore

all the angles at the bases of the triangles are together greater than all the angles of the polygon: But 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 polygon BCDEF; and all the angles of the polygon, together with four right angles, are also equal to twice as many right angles as there are sides in the polygon-therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles: But it has been shewn that all the angles at the bases of the triangles are together greater than all the angles of the polygon; therefore the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, are together less than four right angles. Wherefore, Every solid angle &c.

Q. E. D.