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Book XI.

IF

PROP. XV. THE OR.

F two ftraight lines meeting one another, be parallel see N. to two straight lines which meet one another, but are not in the fame plane with the first two; the plane which paffes through thefe is parallel to the plane paffing through the others.

Let AB, BC two ftraight lines meeting one another, be parallel to DE, EF that meet one another, but are not in the fame plane with AB, BC. the planes through AB, BC, and DE, EF fhall not meet though produced.

3

From the point B draw BG perpendicular to the plane which 2. 11. 11. paffes through DE, EF, and let it meet that plane in G; and

through G draw GH parallel to ED, and GK parallel to EF. b. 31. 1. and because BG is perpendicular to the plane through DE, EF, it

fhall make right angles with e

E

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lines GH, GK in that plane

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both in the fame plane with it) the angles GBA, BGH are to

e

gether equal to two right angles. and BGH is a right angle, e. 29. 1. therefore allo GBA is a right angle, and GB perpendicular to A. for the fame reafon, GB is perpendicular to BC. fince therefore the ftraight line GB ftands at right angles to the two fra aft lines BA, BC, that cut one another in B; GB is perpendicular £ 4. 15. to the plane through BA, BC. and it is perpendicular to the plane through DE, EF; therefore BG is perpendicular to each of the planes through AB, BC and DE, EF. but planes to which the fame ftraight line is perpendicular, are parallel to one anot . g. 14. 11. therefore the plane thro' AB, BC is parallel to the plane thi DE, EF.

Wherefore if two ftraight lines, &c. Q. E. D.

Book XI.

Soc N.

IF

PROP. XVI. THEOR.

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

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

For, if it is not, EF, GH fhall meet, if produced, either on the fide of FH, or EG. firft, let them be produced on the fide of FH, and meet in the point K. therefore fince EFK is in the plane AB,

every point in EFK is in that

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of FH. in the fame manner it may be proved that EF, GH do not meet when produced on the fide of EG. but ftraight lines which are in the fame plane and do not meet, though produced either way, are parallel. therefore EF is parallel to GH. Where fore if two parallel planes, &c. Q. E. D,

IF

PRO P. XVII. THEOR.

F two ftraight lines be cut by parallel planes, they fhall be cut in the fame ratio.

Let the ftraight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D. as AE is to EB, fo is CF to FD.

Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. because the two parallel planes KL, MN are cut by the plane EBDX, the common fections EX, BD are paral,

lel. for the fame reafon, because the two parallel planes GH, KL Book XI.

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Fa ftraight line be at right angles to a plane, every plane which paffes thro' it fhall be at right angles to that plane.

Let the ftraight line AB be at right angles to the plane CK. every plane which paffes through AB fhall be at right angles to the plane CK.

D

Gr A. H

Let any plane DE pass through AB, and let CE be the common festion of the planes DE, CK; take any point F in CE, from which draw FG in the plane DE at right angles to CE. and because AB is perpendicular to the plane CK, therefore it is alfo perpendicular to every ftraight line in that plane meeting it. and confequently it is perpendicular to CE, wherefore ABF is a right angle; but GFB is like

C

K

a. 3. Def.1

FBE

wife a right angle; therefore AB is parallel to FG. and AB is at b 28. 1. right angles to the plane CK; therefore FG is alfo at right angles

to the fame plane. but one plane is at right angles to another plane c. 9. 11. when the ftraight lines drawn in one of the planes, at right angles

Bock XI. to their common fection, are alfo at right angles to the other w plane; and any ftraight line FG in the plane DE, which is at d.4.Def.11. right angles to CE the common fection of the planes, has been

proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all the planes which pafs through AB are at right angles to the plane CK. Therefore if a straight line, &c. Q. E. D.

PROP. XIX. THE OR.

IF two planes cutting one another be each of them perpendicular to a third plane; their common fection fhall be perpendicular to the fame plane.

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

B

If it be not, from the point D draw, in the plane AB, the ftraight line DE at right angles to AD the common fection of the plane AB with the third plane; and in the plane BC draw DF at right angles to CD the common fection of the plane BC with the third plane. and becaufe the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD their common fection, DE is perpendicular to the third a.4.Def.11. plane 2. in the fame manner, it may be

proved that DF is perpendicular to the
third plane. wherefore from the point D
two ftraight lines ftand at right angles to
the third plane, upon the fame fide of it,

EF

D

b. 13.11. which is impoffible b. therefore from the A

C

point D there cannot be any ftraight line at right angles to the

third plane, except BD the common fection of the planes AB,
BC. BD therefore is perpendicular to the third plane.
fore if two planes, &c. Q. E. D.

Where

Book XI.

IF

PROP. XX. THEOR.

a folid angle be contained by three plane angles, See N. any two of them are greater than the third.

Let the folid 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 lefs than either of the other two, and is greater than one of them DAB; and at the point A in the ftraight line AB, make in the plane which paffes through BA,

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AC, the angle BAE equal to the angle DAB; and make AE a. 23. x. equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and jon DB, DC. and because DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, and the angle DAB is equal to the angle

D

A

EAB. therefore the bafe DB is equal b

b. 4. 1.

to the bafe BE. and because BD, DC B

E C

are greater than CB, and one of them BD has been proved equal c. 20. 1. 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 bafe DC greater than the bafe EC; therefore the angle DAC is greater than the angle EAC; and, by the d. 15. x. conftruction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than the angle BAC. but BAC is not lefs than either of the angles DAB, DAC, therefore BAC with either of them is greater than the other. Wherefore if a folid angle, &c. Q E. D.

PROP. XXI. THEOR.

EVERY folid angle

VERY folid angle is contained by plane angles
which together are less than four right angles.

Firft, Let the folid angle at A be contained by three plane angles BAC, CAD, DAB. these three together are lefs than four right angles.

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