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 perpendicular to one and the fame Plane. I fay, A B 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 A B; and join BE, AE, AD.

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Then becaufe A B is at Right Angles to the aforefaid Plane, it fhall be at Right Angles to all Right Def. 3. Lines, touching it, drawn in the Plane; but A Bebis. touches BD, BE, which are in the said Plane; therefore each of the Angles A BD, ABE, is a Right Angle. So, for the fame Reafon, likewife, is each of the Angles CDB, CDE, a Right Angle. Then, becaufe A B is equal to D E, 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 B E.† 4. 1. Again, because AB is equal to D E, and AD to BE; the two Sides A B, BE, are equal to the two Sides ED, DA; but A E, their Bafe, is common; wherefore the Angle ABE is equal to the Angle EDA.† S. 1. But A B E is a Right Angle; therefore EDA is alfo a Right Angle; and fo ED is perpendicular to D A: 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* 5 of this. BD, DA, are in the fame Plane as A B is; for every Triangle is in the fame Plane; therefore it is neceí-† 2 of ibis. fary, that A B, BD, DC, he in one Plane, But both the Angles ABD, BDC, are Right Angles; wherefore AB is parallel to CD. Therefore, ifi 28. 1. two Right Lines be perpendicular to one and the fame Plane, thofe Right Lines are parallel to one another; which was to be demonftrated.

PROPOSITION VII.

THE ORE M.

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

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

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 3 of this. Parallels are, let* be the Right Line EF; then the two Right Lines EGF, EF, will include a Space, Ax. 10. 1. which is + abfurd: Therefore a Right Line, drawn from the Point E to the Point F, is not elevated above the Plane; and, confequently, it must be in that paffing thro' the Parallels A B, CD. Wherefore, if there be two parallel Lines, and any Points be taken in both of them, the Right Line joining thofe 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 Lines, one of which is perpendicular to fome Plane; then fhall the other be perpendicular to the fame Plane.

of Prop. VI. LET AB, CD, be two parallel Right Lines, one of which, as A B, is perpendicular to fome Plane. I fay, the other, CD, is alfo perpendicular to the fame Plane.

For, let AB, CD, meet the Plane in the Points B, D; and let BD be joined; then A B, CD, BD, are *7 of this. * in one Plane. Let DE be drawn in the other

Plane,

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Plane, at Right Angles to BD, and make D E equal to A B; and join BE, A E, AD: Then, fince A B is perpendicular to the Plane, it will be perpendicu-* Def. 3. lar to all Right Lines touching it, that are drawn in the fame Plane; therefore each of the Angles A BD, ABE, is a Right Angle. And fince the Right Line BD falls on the Right Lines A B, CD; the Angles ABD, CD B, shall be + equal to two Right Angles: † 29. 1. Therefore the Angle CDB is also a Right Angle; and fo CD is perpendicular to D B. And fince AB is equal to DE, and BD is common; the two Sides A B, BD, are equal to the two Sides E D, D B. But the Angle A B D is equal to the Angle ED B; for each of them is a Right Angle; therefore the Base AD is equal to the Bafe BE. Again, fince AB ist 4. 1. equal to D'E, and BE to AD; the two Sides A B, BE, fhall be equal to the two Sides E D, DA, each to each: But the Bafe AE is common; wherefore the Angle ABE is equal to the Angle EDA: But 8. 1. the Angle A BE is a Right Angle; therefore E DA 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 tatt 4 of this. Right Angles to all Right Lines, drawn in the faid Def. 3. 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 AB 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 commm 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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PROPOSITION IX.

THEOREM.

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

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ET both the Right Lines A B, CD, be parallel
to the Right Line EF, not being in the fame
Plane with it. I fay, A B is parallel to CD."

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For affume any Point G in EF, from which Point Glet GH be drawn, at Right Angles to E F, in the Plane paffing thro' EF, AB: Alfo let G K be drawn at Right Angles to EF in the Plane paffing thro' E F, CD: Then, because EF is perpendicular to GH and G K, the Line EF fhall alfo be at Right Angles to a Plane paffing thro' both G H and GK: But EF is parallel to AB; therefore A B is + alfo at Right Angles to the Plane paffing thro' HGK. For the fame Reafon, CD is alfo at Right Angles to thë Plane paffing thro' HGK; and therefore A B, and CD, will be both at Right Angles to the Plane pasfing thro' HGK. But if two Right Lines be at 6.of th. Right Angles to the fame Plane, they fhall be * parallel to each other; therefore A B is parallel to CD. And fo, Right Lines that are parallel to the fame Right Line, not being in the fame Plane with it, are alfo parallel to each other; which was to be demonstrated.

PROPOSITION X.

THEOREM.

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

LET 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. I fay, the Angle ABC is equal to the Angle DE F.

For,

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For, take B A, BC, ED, EF, equal one to another, and join AD, CF, BE, AC, DF: Then, because BA is equal and parallel to ED, the Right Line A D fhall also be equal and parallel to BE.* 33. I For the fame Reafon, 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 fame Right Line, not being in the fame Plane with it, will be + parallel to each other. Therefore A D† 9 of this. is parallel and equal to CF; but AC, DF, join them; wherefore A C is equal and parallel to DF.1 23. 1. And because the two Right Lines A B, BC, are equal to the two Right Lines DE, E F, and the Bafe A C equal to the Bafe DF; therefore the Angle ABC will be equal to the Angle DEF. Whence, if* 8. two Right Lines, touching one another, be parallel to two other Right Lines touching one another, but not in the Jame Plane, thofe Right Lines contain equal Angles; which was to be demonstrated.

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

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ETA be the Point given, above the given Plane 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,* 12. 1. perpendicular to BC; then if AD be perpendicular to the Plane BH, 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 G H, parallel to B C.

Then, because BC is perpendicular to both D A and DE, BC will also be + perpendicular to a Plane † 4 of this. 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 the other beat Right Angles to the fame Plane : † 8 of this. Wherefore GH is at Right Angles to the Plane paf