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.

PROPOSITION XIX.

Q. B. D.

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 first two; BD is perpendicular to the third plane.

If it be 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 in the plane BC draw DF at right angles to CD, the common section of the plane BC with the third plane.

And 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 per• 4 Def. 11. pendicular to the third plane*:

+ 13. 11.

In the same manner it may be proved, that DF is perpendicular to the third plane;

C

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 t; 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; BD therefore is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D.

PROPOSITION 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.

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, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and 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; and make AE equal to 23. 1. AD, and through E draw BEC cutting AB, AC, in the points B, C, and join

DB, DC.

And because DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, each to each; and the angle DAB is equal to the angle EAB; therefore the base DB is equal to the baseBE.

+ 4.1.

And because BD, DC are greater than CB, and one of 20. 1. them BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC:

And because DA is equal to AE, and AC common, but the base DC greater than the base EC; therefore the angle DAC is greater than the angle EAC; and, by the con- || 25.1. struction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle 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. Q. E. D.

PROPOSITION XXI.

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 three together are less than four right angles.

Take, in each of the straight lines AB, AC, AD, 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 them are greater than the third; therefore the angles CBA, ABD 20. 11. are greater than the angle DBC:

+ 32. 1.

20. 11.

For the same reason, the angles BCA, ACD are greater than the angle DCB; and the angles CDA, ADB greater than BDC; wherefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles DBC, BCD, CDB:

But the three angles DBC, BCD, CDB are equal to two right anglest; therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater

than two right angles:

And because the three an-
gles of each of the triangles
ABC, ACD, ADB are equal
to two right angles, therefore
the nine angles of these three
triangles, viz. the angles CBA,
BAC, ACB, ACD, CDA,
DAC, ADB, DBA, BAD
are equal to six right angles: B

C

Of these, the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles; therefore the remaining three angles BAC, DAC, BAD, which contain the solid angle at A, are 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 together are 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:

And because the solid an-
gle at B is contained by three
plane angles CBA, ABF,
FBC, of which any two are
greater than the third, the
angles CBA, ABF are greater B
than the angle 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

F

bases of the triangles are together greater than all the angles of the polygon :

And because all the angles of the triangles are together equal to twice as many right angles as there are triangles||, | 32. 1. that is, as there are sides in the polygon BCDEF; and that all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon§; therefore all the angles of § 1 Cor. 32.1. the triangles are equal to all the angles of the polygon together with four right angles: but all the angles at the bases of the triangles are greater than all the angles of the polygon, 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. Q. E. D.

THE END.

Printed by Metcalfe and Palmer, Trinity-Street, Cambridge.