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PROPOSITION IX.

THE ORI M.
Right Lines that are parallel to the same Right

Line, not being in the fame Plane with it, are
also parallel to each other.
ET both the Right Lines AB, CD, be parallel

to the Right Line EF, not being in the same Plane with it. I fay, AB is parallel to CD.

For affume any Point Gin EF, from which Point G, let GH be drawn at Right Angles to E F, in the Plane paffing thro' EF, AB: Also let GK be drawn at Right Angles to E F in the Plane passing thro” EF, CD: Then because EF is perpendicular to GH, and GK, the Line E F shall also be * at Right Angles to a Plane * 4 of this. paffing thro' GH, GK; but EF is parallel to AB. Therefore AB is also at Right Angles to the Planet 8 of tbisa passing thro' HGK. For the same Reason, C D is also åt Right Angles to the Plane passing thro' HGK; and therefore AB and CD, will be both at Right Angles to the Plane passing thro' HGK. But if two Right Lines be at Right Angles to the fame Plane, they shall be f parallel to each other. Therefore AB is 1.6 of this. parallel to CD; which was to be demonstrated.

PROPOSITION X.

THEOREM.
If two Right Lines, touching one another, be pa-

rallel to two other Right Lines, touching one
another, but not in the same Plane, these Right
Lines contain equal Angles.
ET two Right Lines A B, BC, touching one

another, be parallel to two Right Lines DE, EF, touching one another, but not in the fame Plane. Í fay, the Angle ABC is equal to the Angle DEF.

For take BA, BC, ED, EF, equal to one another, and join AD, CF, BE, AC, DF: Then because BA is equal and parallel to ED, the Right Line

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30. I.

AD shall also be * equal and parallel to BE. For the fame Reason, CF will be equal and parallel to BE; therefore AD, CF, are both equal and parallel to BE. But Right Lines that are parallel to the same

Right Line, not being in the fame Plane with it, will + 9 of this. be + parallel to each other. Therefore AD is parafı

lel and equal to CF, but AC, DF, joins them; 33. 1. wherefore AC is equal and parallel to DF. And

because two Right Lines AB, BC, are equal to two Right Lines DE, EF, and the Base AC equal to the Base DF, the Angle ABC will be * equal to the Angle DEF. Therefore, if two Right Lines, touching one another, be parallel to two other Right Lines, touching one another, but not in the same Plane; those Right Lines contain equal Angles; which was to be demonstrated.

* 8. I.

PROPOSITION XI.

PROBLEM,
From a Point given ałote a Plarie, to draw a Right

Line perpendicular to that Plane.

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12. 1.

ET A be a Point given above the given Plane

BH. It is required to draw a Right Line from the Point A, perpendicular to the Plane BH.

Let a Right Line B C be any how drawn in the Plane BH, and let AD be drawn* from the Point A perpendicular to BC; then if AD be perpendicular to the Plane BH, the thing required is already done. But if not, let DÉ be drawn in the Plane from the Point D at Right Angles to BC; and let AF be drawn * from the Point A perpendicular to DE. Lastly, thro' F draw GH parallel to BC

Then because B C is perpendicular to both DA and * 4 of sbis. DE, BC will also be + perpendicular to a Plane paf

fing thro' ED, DA. But GH is parallel to BC. And if there are two Right Lines parallel, one of

which is at Right Angles to fome Plane, then thall I 8 of this. the other be I at Right Angles to the fame Plane.

Wherefore G H is at Right Angles to the Plane paffing * Def. 3. thro' ED, DA, and so is * perpendicular to all the Right Lines in the fame Plane that touch it. But AF,

which

which is in the Plane passing thro' ED and DA, doth touch it. Therefore GH is perpendicular to AF, and so AF is perpendicular to GH; but AF likes wise is perpendicular to DE; therefore AF is perpendicular to both HG, DE. But if a Right Line stands at Right Angles to two Right Lines, in their common Section, that Line will be † at Right An- + 4 of tbis. gles to the Plane passing thro’ these Lines. Therefore AF is perpendicular to the Plane drawn thro ED, GH; that is, to the given Plane BH. Therefore AF is drawn from the given Point A, above the given Plane BH, perpendicular to the said Plane ; which was to be done.

PROPOSITION XII.

.

PROBLEM.
To ere&t a Right Line

perpendicular to a given
Plane, from a Point given therein.
LE
ET A be a given Point in a given Plane MN.

It is required to draw a Right Line from the
Point A, at Right Angles, to the Plane MN.

Let some point B be supposed above the given Plane, from which let B C be drawn * perpendicular * 11 of this to the said Plane; and let AD be drawn † from A 731. I. parallel to BC.

Then because AD, CB, are two parallel Right Lines, one of which, viz. BC, is perpendicular to the Plane MN; the other AD shall be I also perpendi- | 8 of this cular to the fame Plane. Therefore, a Right Line is erected perpendicular to a given Plane, from a Point given therein; which was to be done.

PROPOSITION XIII.

THEOR E M.
Two Right Lines cannct be erected at Right Angles,

to a given Plane from a Point (herein given.
FAR

OR, if it is poffible, let two Right Lincs AB,

AC, be erected perpendicular to a given Plane on the same Side, at a given Point A, in a given Plane.

Then let a Plane be drawn thro' BA, AC, cutting * 3 of tbis. the given Plane thro' A in the Right Line * DAE;

therefore the Right Lines AB, AC, DAE, are in one

Plane. And because CA iş perpendicular to the + Def. 3. given Plane, it shall also be † perpendicular to all

Right Lines drawn in that Plạne, and touching it ; but DAE being in the given Plane, touches it. Therefore the Angle CAE is a Right Angle. For the same Reason, B.E is also a Right Angle; wherefore the Angle CAE is equal to BAE, and they are both in one Plane, which is absurd. Therefore, two Right Lines, cannot be erected at Right Angles, to a given Plane, from a Point therein given ;" which was to be demonstrated.

PROPOSITION XIV.

THEOREM.
Thoje Planes, to which the same Right Line is per.

Tendicular, are parallel to each other.
I

of the Planes CD, EF. I fay, these Planes arc parallel

For if they be not, let them be produced till they meet each other, and let the Right Line GH be the common Section, in which take any Point K, and join AK, BK. Then because AB is perpendicular to the Plane EF, it shall also be perpendicular to the Right Line BK, being in the Plane E F produced. Wherefore the Angle ABK is a Right Angle. And for the same Reason, BAK is allo a Right Angle.

And

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And so the two Angles ABK, BAK, of the Trian. gle ABK, are equal to two Right Angles, which is

impoffible. Therefore the Planes CĎ, EF, being * 19. 1, produced, will not meet each other, and so are neceffarily parallel. Therefore, those Planes, to which the fame Right Line is perpendicular, are parallel to each other; which was to be demonstrated.

PROPOSITION XV,

THEOREM,
If two Right Lines, touching one another, be pa-

rallel to two Right Lines, touching one anotber,
and not being in the same Plane with them, the
Planes drawn thro' those Right Lines are pa-
rallel to each otber.

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ET two Right Lines AB, BC, touching one touching one another, but not in the same Plane with them. I say, the Planes passing thro' AB, BC, and DE, EF, being produced, will not meet each other.

For let BG be drawn from the Point B, perpendicular to the Planè paffing thro' DE, EF, meeting the same in the Point G; and thro' G let GH be drawn parallel to ED, and GK parallel to EF; then because BG is perpendicular to the Plane passing thro'. DE, EF, it shall also make * Right Angles * Def. 3. with all Right Lines that touch it, and are in the fame Plane; but GH and GK, which are both in the fame Plane, touch it. Therefore each of the Angles BGH, BGK, is a Right Angle. And since BĂ is parallel to GH, the Angles GBA, GBH, are ++ 29. r. equal to the Right Angles : But BGH is a Right Angle; wherefore GBA fhall also be a Right Angle, and lo BG is perpendicular to BA. For the same Reason, GB is also perpendicular to BC. Therefore since a Right Line BG, stands at Right Angles to two Right Lines BA, BC, mutually cutting each other; BG Ihall also be f at Right Angles to the Plane drawnt 4 of tóls. thro' BA, B.C. But it is perpendicular to the Plane drawn thro' DE, EF; therefore BG is perpendicujar to both the Plançs drawn thro' AB, BC, and

DE,

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