Book XI.

a 33. I.

b 9.

€ 8. 1.

à 12. I.

b 11. 1.

C 31. I.

Let two right lines AB, BC, touching one another, be paral lel to two right lines DE, EF, touching one another, but not in the fame plain, the angle ABC is equal to the angle DEF. For, take BA, BC, ED, EF, equal to one another, and join AD, CF, EB, AC, DF; then, because AB, DE, are equal and parallel, AD, BE, that join them, are likewife equal and parallel. For the fame reafon, CF, BE, are equal and parallel; then AD, CF, are equal and parallel ; therefore AC, DF, that join them, are equal and parallel ; then, fince AB, BC, are equal to DE, EF, and the bafes AC, DF, are equal, the angle ABC is equal to DEF. Wherefore, &c.

T

PRO P. XI. PRO B.

O let fall a perpendicular on a given plain from a given point above it.

Let BH be the given plain, and A the point above it; it is required from the point A to let fall a perpendicular upon the given plain BH. In the plain BH take any right line BC; and from the point A draw AD perpendicular to BC. If AD is perpendicular to BH, what was required is done; if not, draw DE in the plain at right angles to BC; and from A draw AF perpendicular to DE ; and through F draw GH parallel to BC. Then, becaufe BC is perpendicular both to DA and DE, it is d def. 3. perpendicular to the plain paffing through DA, DE 4; but GH is parallel to BC; therefore GH is perpendicular to the plain pafling through DA, DE; therefore AF is perpendicular to GH ; but AF is perpendicular to DE; therefore at right angles to the plain paffing through GH, ED; that is, to BH. Wherefore, &c.

€ 7.

f def. 3.

84.

T

PRO P. XII. PRO B.

O erect a right line perpendicular to a given plain from a given point in it.

It is required to draw a perpendicular to the plain MN, from a given point A in it. From fome point B, above the plain, let fall a perpendicular BC upon it; and from the point A draw AD parallel to BC b. Then, because AD, BC, are two paral lel right lines, BC, one of them, is perpendicular to the plain MN; the other, AD, is perpendicular to the fame plain. Wherefore, &c.

T

PROP. XIII. THEOR.

WO right lines cannot be drawn at right angles to a given
plain, from a point given therein.

For, if poffible, let the right lines AB, AC, be drawn perpen, dicular to a given plain, from the given point A; let a plain paffing through AB, AC, cutting the given plain through A, in the right line DAE; but the right line DAE being in the given plain touches it; therefore AB, AC, DAE, are in one plain; then, because AC is perpendicular to the given plain, the angle CAE is a right angle b; for the fame reason BAE is a right angle; therefore the angle BAE is equal to the angle CAE, a part to the whole, which is abfurd. Wherefore, &c.

T

PROP. XIV. THE OR.

HOSE plains to which the fame right line is perpendicu-
lar, are parallel to each other.

a

Book XI.

a 3.

b def. 3.

If the right line AB be perpendicular to each of the plains DC, EF; then these plains are parallel: For, if not, let them be produced till they meet each other; and let the right line GH be their common fection; in which, take any point K, and join AK, BK; then, because AB is perpendicular to the plain EF, it is perpendicular to the right line BK, being in a def. 3. the fame plain produced; therefore ABK is a right angle; for the fame reason BAK is a right angle; that is, two angles in a triangle equal to two right angles, which cannot be; therefore the plains CD, EF, being produced, will not meet each other; therefore parallel. Wherefore, &c.

IF F two right lines, touching one another, be parallel to two other right lines, touching one another, and not in the fame plain with them, the plains drawn through these right lines are parallel to each other.

Let AB, BC, two right lines touching one another, be parallel to two right lines DE, EF, touching one another, but not in the fame plain with them; then the plains paffing through AB, BC, DE, EF, being produced, will not meet each other: For, from the point B, draw the right line BG2, to the point a 11.

G;

b 31. 1.

c def. 3.

Book XI. G, in the plain paffing through DE, EF, and perpendicular to it. From the point G, draw GH parallel to DE, and GK parallel to EF; then, becaufe BG is perpendicular to the plain paffing through DE, EF, it is perpendicular to all the right lines touching it in that plain; therefore BG is perpendicular to GH, GK; and, fince BA is parallel to GH, the angles GBA, BGH, are equal to two right angles; but BGH is a right angle; therefore ABG is likewife a right angle: For the fame reason, GBC is a right angle; therefore BG is at right angles to the plain paffing through BA, BC; but it is likewife perpendicular to the plain paffing through DE, EF; therefore BG is perpendicular to the plains paffing through BA, BC, and ED, EF: Therefore these plains are parallel. Wherefore,

d 29. 1.

&c.

PRO P. XVI. THE OR.

IF two parallel plains are cut by another plain, their common fections will be parallel.

Let the two parallel plains AB, CD, be cut by any plain EFGH, whofe common fections are EF, GH; then EF is parallel to GH: For, if EF, GH, are not parallel, if produced, they will meet, either toward F, H, or E, G. Let them meet in K: Then, because EFK is in the plain AB, all points taken in it are in the fame plain; therefore K is in the plain AB: For the fame reason, K is in the plain CD; therefore the plains AB, CD, meet each other; but they are parallel; therefore they cannot meet; for the fame reafon they cannot meet, if produced toward E, G; therefore the common fections EF, GH, are parallel. Wherefore, &c.

PRO P. XVII. THE OR.

IF two right lines are cut by parallel plains, they will be cut in the fame proportion.

Let the two right lines AB, CD, be cut by parallel plains GH, KL, MN, in the points A, E, B, C, F, D; then AE will be to EB as CF is to FD: For, let AC, BD, AD, be joined; and let AD meet the plain KL, in the point X, join EX, XF; then, becaufe the plains KL, MN, are cut by the

plain EBDX, their common fections EX, BD, are parallel ";Book VI. for the fame reason, the common fections XF, AC, are parallel then, fince EX is parallel to BD, AE is to EB as AX is a 16. to XD; for the fame reafon AX is to XD as CF is to FD; therefore AE is to EB as CF is to FD. Wherefore, &c.

PRO P. XVIII. THE OR.

IF a right line is perpendicular to fome plain, then all plains paffing through that line will be perpendicular to the fame plain.

b 2.6.

Let the right line AB, be perpendicular to the plain CL; then all plains paffing through AB are perpendicular to the fame plain. For, let a plain DE pafs through the right line AB, whofe common fection with the plain CL is the right line CE; from fome point F, in CE, draw FG in the plain DE, perpendicular to the right line CE; then, because AB is perpendicular to the plain CL, it is perpendicular to the right line CE in it; therefore the angle ABF is a right angle; but a def. 3. GFB is likewife a right angle; therefore AB is parallel to FG; but AB is at right angles to the plain CL; therefore FG is at right angles to the fame plain: For the fame reason, ¢ 7. all other lines drawn perpendicular to the common fection of any plain paffing through AB, is perpendicular to CL: Then, because AB, FG, are drawn in one plain, perpendicular to CE, the common section of the plains CH, CL, the plain CH is perpendicular to the plain CLa. Wherefore, &c.

IF

PRO P. XIX. THE OR.

F two plains cutting each other, be perpendicular to fome plain, their common fection will be perpendicular to that same plain.

b 28. I.

d def. 4

Let two plains AB, BC, cutting each other, be perpendicu lar to fome third plain, ADC; their common fection BD is perpendicular to the plain ADC: For, if BD is not perpendicular to ADC, from the point D, draw DE in the plain AB, a def. 4. at right angles to AD; and DF in the plain CB, at right angles to DC; then the two right lines DE, DF, are drawn from the fame point, each at right angles to the fame plain ADC; which is impoffible; therefore BD is perpendicular 13. to ADC. Wherefore, &c.

PROP.

Book XI.

a 23. I.

b constr.

C 4. I.

d 20. I.

e 25. I.

PRO P. XX. THE OR.

IF a folid angle be contained under three plain angles, any two

of them, however taken, are greater than the third.

Let the folid angle A, be contained under three plain angles BAC, CAD, BAD, any two of them, however taken, are greater than the third.

If the three angles, or any two of them, are equal, then any two of them must be greater than the third; but, if not equal, let one of them, as BAC, be the greater: At the point A, make the angle BAE, with the right line AB, in the plain paffing through BA, AC, equal to the angle DAB2; make AE equal to AD; through 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, the two fides BA, AD, are equal to the two fides BA, AE; and the angle BAD equal to the angle BAE; then the bafes BD, BE, are equal; but the two fides BD, DC, are greater than the third BC; and BD is equal to BE; therefore the remainder DC, is greater than EC; and the fides CA, AD, are equal to the two fides CA, AE, and the bafe DC greater than CE; therefore the angle DAC is greater than EAC; but the angles BAE, EAC, are equal to BAC, therefore the angles BAĎ, DAC, are greater than BAC. After the fame manner, any other two angles may be proved greater than the third. Wherefore, &c.

8 2.

E

PROP. XXI.

VERY folid angle is contained under plain angles, together lefs than four right angles.

Let the angle A be a folid angle, contained under the plain angles BAC, BAD, DAC; thefe angles are lefs than four right angles. For, in the lines, AB, AD, AC, take any points B, D, C, and join BD, DC, BC; then, because the folid angle. at B is contained under three plain angles, CBA, ABD, DBC, any two of which are greater than the third; the two angles, CBA, ABD, are greater than DBC; for the fame reason, the angles BCA, ACD, are greater than BCD; and CDA, ADB, are greater than BDC: Therefore the fix angles ABD, ABC, ACE, ACD, ADC, ADB, are greater than the three angles