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14. 1.

PROPOSITION VI.

THEOREM.

If two Right Lines be perpendicular to one and the fame Plane, thofe Right Lines are parallel to one another.

LET two Right Lines AB, CD, be perpendicu

lar to one and the fame Plane. I fay, AB is parallel to CD.

For let them meet the Plane in the Points B, D, and join the Right Line BD, to which let DE be drawn in the fame Plane at Right Angles; make DE equal to AB, and join BE, AE, AD. -

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Then becaufe AB is at Right Angles to the afore* Def. 3. f faid Plane, it fhall be at Right Angles to all Right Lines, touching it, drawn in the Plane; but A B touches BD, BE, which are in the faid Plane. Therefore each of the Angles ABD, ABE, is a Right Angle. So for the fame Reafon likewife, is each of the Angles CDB, CDE, a Right Angle. Then because AB is equal to DE, and BD is common, the two Sides AB, BD, fhall be equal to the two Sides ED, DB; but they contain Right Angles. Therefore the Bafe AD is equal to the Bafe BE. Again, because AB is equal to DE, and AD to BE, the two Sides AB, BE, are equal to the two Sides ED, DA; but AE, their Base, is common. Wherefore the Angle ABE is equal to the Angle EDA; but ABE is a Right Angle. Therefore EDA is alfo a Right Angle; and fo ED is perpendicular to DA; but it is alfo perpendicular to BD and DC. Therefore ED is at Right Angles in the Point of Contact to three Right Lines BD, DA, DC. Wherefore these three laft Right Lines are * in one Plane: But BD, DA, are in the fame Plane as AB is; for every Triangle is † in the fame Plane. Therefore it is neceffary that AB, BD, DC, be in one Plane; but both the Angles ABD, BDC, are Right Angles. Wherefore AB is parallel to CD. Therefore, if two Right Lines be perpendi cular to one and the fame Plane, thofe Right Lines are parallel to one another; which was to be demonftrated.

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PRO

PROPOSITION VII.

THEORE M.

If there be two Parallel Lines, and any Point be taken in both of them, the Right Lines joining thofe Points fhall be in the fame Planes as the Parallels are.

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ET AB, CD, be two parallel Right Lines, in which are taken any Points E, F. I fay, a Right Line joining the Points E, F, are in the fame Plane as the Parallels are.

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For if it be not, let it be elevated above the fame, if poffible, as EGF; thro' which let fome Plane be drawn, whofe Section, with the Plane in which the Parallels are, let be the Right Line EF, then the* 3 of thise two Right Lines EGF, EF, will include a Space, which is abfurd. Therefore a Right Line drawn † Axiom from the Point E to the Point F, is not elevated 10. I. above the Plane, and confequently it must be in that paffing thro' the Parallels AB, CD. Wherefore, if there be two parallel Lines, and any Points be taken in both of them, the Right Line_joining thefe Points fhall be in the fame Plane as the Parallels are; which

was to be demonftrated.

PROPOSITION VIII.

THEOREM.

If there be two parallel Right Linès, one of which is perpendicular to fome Plane, then shall the other be perpendicular to the fame Plane.

ET AB, CD, be two parallel Right Lines, one See the Fig. of which, as AB is perpendicular to fome of Prop. VI. Plane. I fay, the other CD is also perpendicular to the fame Plane.

For let AB, CD, meet the Plane in the Points B, and let BD be joined; then AB, CD, BD, are

in one Plane, Let DE be drawn in the Plane at * 7 of this. Right Angles to BD, and make DE equal to AB,

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and join BE, AE, AD. Then fince AB is perpendicular to the Plane, it will be perpendicular to all Right Lines, touching it, drawn in the fame Plane; therefore each of the Angles ABD, ABE, is a Right Angle. And fince the Right Line BD falls on the Right Lines AB, CD, the Angles A BD, CDB fhall be equal to two Right Angles. Therefore the Angle CDB is alfo a Right Angle, and fo CD is perpendicular to DB: And fince A B is equal to DE, and BD is common, the two Sides AB, BD, are equal to the two Sides ED, DB. But the Angle ABD is equal to the Angle EDB; for each of them is a Right Angle. Therefore the Bafe AD is equal to the Bafe BE. Again, fince A B is equal to DE, and BE to AD, the two Sides AB, BE, fhall be equal to the two Sides ED, DA, each to each; but the Bafe A E is common. Wherefore the Angle ABE *8. is equal to the Angle EDA; but the Angle ABE is a Right Angle. Therefore EDA is alfo a Right Angle, and ED is perpendicular to DA; but it is likewife perpendicular to DB: Therefore ED fhall alfo be † perpendicular to the Plane paffing thro' BD, DA, and likewife fhall be at Right Angles to all Right Lines, drawn in the faid Plane that touch it. But DC is in the Plane paffing thro' BD, DA, because AB, BD, are in that Plane; and DC is + in the fame Plane that A B and BD are in. Wherefore ED is at Right Angles to DC, and fo CD is at Right Angles to DE, as alfo to DB. Therefore CD ftands at Right Angles in the common Section D, to two Right Lines DE, DB, mutually cutting one another; and accordingly is at Right Angles to the Plane paffing thro' DE, DB; which was to be demonftrated.

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PRO

PROPOSITION IX.

THEORE M.

Right Lines that are parallel to the fame Right Line, not being in the fame Plane with it, are alfo parallel to each other.

LET both the Right Lines AB, CD, be paralle!

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

For affume any Point G in EF, from which Point G, let GH be drawn at Right Angles to EF, in the Plane paffing thro' EF, AB: Alfo let GK be drawn at Right Angles to EF in the Plane paffing thro' EF, CD: Then because EF is perpendicular to GH, and GK, the Line EF fhall alfo be at Right Angles to a Plane 4 of this. paffing thro' GH, GK; but EF is parallel to AB. Therefore AB is† alfo at Right Angles to the Plane † 8 of this, paffing thro' HGK. For the fame Reason, CD is also at Right Angles to the Plane paffing thro' HGK; and therefore A B and CD, will be both at Right Angles to the Plane paffing thro' HGK. But if two Right Lines be at Right Angles to the fame Plane, they fhall be parallel to each other. Therefore AB is 1.6 of this. parallel to CD; which was to be demonftrated.

PROPOSITION X.

THEOREM.

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If two Right Lines, touching one another, be parallel to two other Right Lines, touching one another, but not in the fame Plane, these Right Lines contain equal Angles."

ET two Right Lines AB, BC, touching one

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touching one another, but not in the fame Plane. I 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 becaufe BA is equal and parallel to ED, the Right Line

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AD fhall 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 But Right Lines that are parallel to the fame Right Line, not being in the fame Plane with it, will +9 of this. bet parallel to each other. Therefore AD is paral lel and equal to CF, but AC, DF, joins them; 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 Bafe AC equal to the Bafe 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 fame Plane, thofe Right Lines contain equal Angles; which was to be demonftrated.

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From a Point given above a Plane, to draw a Right
Line perpendicular to that Plane.

ET A be a Point given above the given Plane

LETA a

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

Let a Right Line BC 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 B H, the thing required is already done. But if not, let DE 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, Laftly, thro' F draw GH parallel to B C

Then because BC is perpendicular to both DA and † 4 of this. DE, BC will alfo be † perpendicular to a Plane paffing 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 fhall 8 of this. the other be at Right Angles to the fame Plane. Wherefore G H is at Right Angles to the Plane paffing thro' ED, DA, and fo is perpendicular to all the Right Lines in the fame Plane that touch it. But AF,

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