which let F G be drawn in the Plane DE, perpendicular to the Right Line CE. Then because AB is Def. 3. perpendicular to the Plane CL, it fhall also be * perpendicular to all the Right Lines which touch it, and are in the fame Plane. Wherefore it is perpendicular to CE; and confequently the Angle ABF is a Right Angle; but GFB is likewise a Right Angle. Therefore AB is parallel to F G. But AB is at Right +of this. Angles to the Plane CL. Therefore FG will be + at Right Angles to that fame Plane. But one Plane is perpendicular to another, when the Right Lines, drawn in one of the Planes perpendicular to the common Section of the Planes, are ‡ perpendicular to the other Plane. But FG is drawn in one Plane DE, perpendicular to the common Section CE of the Planes. And it has been proved to be perpendicular to the Plane CL. Therefore the Plane DE is at Right Angles to the Plane CL. After the fame manner it is demonftrated, that all Planes, paffing thro' the Right Line AB, are perpendicular to the Plane CL, Therefore, if a Right Line be perpendicular to fome Plane, then all Planes paffing thro' that Line will be perpendicular to the fame Plane; which was to be demonstrated.

Def 4

of this.

PROPOSITION XIX.

THEORE M.

If two Planes, cutting each other, be perpendicular to fome Plane, then their common Section will be perpendicular to that fame Plane.

ET. two Planes AB, BC, cutting each other, be perpendicular to fome third Plane, and let their common Section be BD. I fay, BD is perpendicular to the faid third Plane, which let be ADC.

For, if poffible, let BD not be perpendicular to the third Plane; and from the Point D, let DE be drawn in the Plane AB, perpendicular to AD; and let DF be drawn in the Plane BC, perpendicular to CD; then because the Plane AB is perpendicular to the third Plane, and DE is drawn in the Plane A B perpendicular to their common Section AD, DE

fhall

Def. 4•

fhall be perpendicular to the third Plane. In like manner we prove, that DF alfo is perpendicular to the faid Plane. Wherefore two Right Lines ftand at Right Angles, to this third Plane, on the fame Side at the fame Point D, which is † abfurd. Therefore tot 13 of this this third Plane cannot be erected any Right Lines, perpendicular at D, and on the fame Side, except BD, the common Section of the Planes AB, BC. Wherefore DB is perpendicular to the third Plane. If, therefore, two Planes, cutting each other, be perpendicular to fome, Plane, then their common Section will be perpendicular to that fame Plane; which was to be demonftrated,

PROPOSITION XX.

THEORE M.

If a folid Angle be contained under three plane An gles, any two of them, howsoever taken, are greater than the third.

LE

H

ET the folid Angle A be contained under three plane Angles BAC, CAD, DAB. I fay any two of the Angles BAC, CAD, DAB, are greater. than the third, howfoever taken.

For if the Angles BAC, CAD, DAB, be equal, it is evident that any two, howfoever taken, are greater than the third. But if not, let BAC be the greater; and make the Angle B AE, at the Point A, * 23. 1. with the Right Line AB, in a Plane paffing thro' BA, AC, equal to the Angle DAB, make AE equal to AD; thro' E draw BEC, cutting the Right Lines, AB, AC, in the Points B, C, and join DB, DC. Then because DA is equal to AE, and AB is common, the two Sides DA, AB, are equal to the two Sides AE, AB; but the Angle DAB is equal to the Angle BAE. Therefore the Base DB ist equal to + 4. I the Bafe BE. And fince the two Sides DB, DC, are greater than BC, and DB has been proved equal to BE, the remaining Side DC fhall be greater than the remaining Side EC; and fince DA is equal to AE, and AC is common, and the Base DC greater than the Bafe EC, the Angle DAC fhall be greatert 25.4

than

T

than the Angle EAC. But from Conftruction, the Angle DAB is equal to the Angle BAE. Wherefore the Angles DAB, DAC, are greater than the Angle BAC After this manner we demonftrate, if any two other Angles be taken, that they are greater than the third. Therefore, if a folid Angle be contained under three plain Angles, any two of them, howsoever taken, are greater than the third; which was to be demonftrated.

PROPOSITION XXI.

THEOREM.

Every folid Angle is contained under plane Angles together less than four Right ones.

LET A

a

ET A be a folid Angle, contained under plane Angles BAC, CAD, DAB. I fay the Angles BAC, CAD, DAB, are lefs than four Right Angles.

*

For take any Points B, C, D, in each of the Lines AB, AC, AD, and join BC, CD, DB. Then because the folid Angle at B is contained under three plane Angles CBA, ABD, CBD, any two of these 20 of this are greater than the third. Therefore the Angles CBA, ABD, are greater than the Angle CBD. For the fame Reason, the Angles BCA, ACD, are greater than the Angle BCD; and the Angle CDA, ADB, greater than the Angle CDB. Wherefore the fix Angles CBA, ABD, BCA, ACD, ADC, ADB, are greater than the three Angles CBD, BCD, CDB. But the three Angles CBD, BCD, CDB, are equal to two Right Angles. Wherefore the fix Angles CBA, ABD, BCA, ACD, ADC, ADB, are greater than two Right Angles. And fince the three Angles of each of the Triangles ABC, ACD, ADB, are equal to two Right Angles, the nine Angles of thofe Triangles CBA, BCA, BAC, ACD, CAD, ADC, ADB, ABD, DAB, are equal to fix Right Angles. Six of which Angles CBA, BCA, ACD, ADC, ADB, ABD, are greater than two Right Angles. Therefore the three other Angles BAC, CAD, DAB, which contain the folid An

† 32. I.

gle,

gle, will be lefs than four Right Angles. Wherefore every folid Angle is contained under Angles together, less than four plane right ones; which was to be demonftrated.

PROPOSITION XXII.

THE ORE M.

If there be three plane Angles, whereof two, any bow taken, are greater than the third, and the Right Lines that contain them be equal, then it is poffible to make a Triangle of the Right Lines joining the equal Right Lines, which form the Angles.

ET ABC, DEF, GHK, be given plane Angles, any two whereof are greater than the third; and let the equal Right Lines AB, BC, DE, EF, GH, HK, contain them; and let AC, DF, GK, be joined. I fay, it is poffible to make a Triangle of AC, DF, GK, that is, any two of them, howfoever taken, are greater than the third.

*

For if the Angles at B, E, H, are equal, then AC, DF, GK, will be equal, and any two of them 4. I. greater than the third; but if not, let the Angles at B, E, H, be unequal, and let the Angle B be greater than either of the others at E or H. Then the Right Line AC will be + greater than either DF or GK; † 244 & † and it is manifest, that AC, together with either DF, or GK, is greater than the other. I fay likewife, that DF, GK, together, are greater than AC. For make tatt 23. 4. ¿ ↑ the Point B, with the Right Line AB, the Angle ABL, equal to the Angle GHK, and make BL equal to either AB, BC, DE, EF, GH, HK, and join AL, CL. Then, because the two Sides AB, BL, are equal to the two Sides GH, HK, each to each, and they contain equal Angles, the Bafe AL fhall be equal to the Bafe GK. And fince the Angles E and Hare greater than the Angle ABC, whereof the Angle GHK is equal to the Angle ABL, the other Angle at E fhall be greater than the Angle LBC, And fince the two Sides LB, BC, are equal to the two Sides DE, EF, each to each, and the Angle

# 24. I.

*

DEF is greater than the Angle LBC, the Bafe DF fhall be greater than the Bafe LC. But GK has been proved equal to AL. Therefore DF, GK are greater than AL, LC; but AL, LC, are greater than AC. Wherefore DF, GK, fhall be much greater than AC. Therefore any two of the Right Lines AC, DF, GK, howfoever taken, are greater than the other: And fo a Triangle may be made of AC, DF, GK; which was to be demonftrated.

PROPOSITION XXIII.

PROBLEM.

To make a folid Angle of three plane Angles, whereof any two, bowfoever taken, are greater than the third; but these three Angles must be less than four Right Angles.

ET ABC, DEF, GHK, be three plane Angles given, whereof any two, howsoever taken, are greater than the other, and let the faid three Angles be less than four Right Angles. It is required to make a folid Angle of three plane Angles equal to ABC, DEF, GHK.

Let the Right Lines AB, BC, DE, EF, GH, HK, be cut off equal, and join AC, DF, GK; then it is 22 of this. poffible to make a Triangle of three Right Lines † 2z. 1. equal to AC, DF, GK: And fo let + the Triangle LMN be made, fo that AC be equal to L M, and DF to MN, and GK to LN; and let the Circle LMN be defcribed about the Triangle, whose Center let be X, which will be either within the Triangle LMN, or on one Side thereof, or without the fame.

15.4

8. I

First, let it be within, and join LX, MX, NX. I fay AB is greater than LX. For if this be not fo, AB fhall be either equal to LX, or lefs. First, let it be equal; then because AB is equal to LX, and also to BC, LX fhall be equal to BC; but LX is equal to XM. Therefore the two Sides AB, BC, are equal to the two Sides LX, XM, each to each; but the Bafe AC is put equal to the Bafe LM. Wherefore the Angle ABC shall be * equal to the Angle