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For take the Right Lines EA, EB, CE, DE, equal, and thro' E any how draw the Right Line GEH, and join AD, CB; and from the Point F let there be drawn FA, FG, FD, FC, FH, FB: Then because two Right Lines AE, ED, are equal to two Right Lines CE, EB, and they contain* the equal Angles AED, CEB; the Bafe AD fhall be t equal to the Bafe CB, and the Triangle AED equal to the Triangle CEB; and fo likewife is the Angle DAE equal to the Angle EBC; but the Angle A EG is equal to the Angle BE H; therefore A GE, BEH, are two Triangles, having two Angles of the one equal to two Angles of the other, each to each, and one Side A E equal to one Side EB, viz. those that are at the equal Angles; and fo the other Sides of the one, will be + equal to the other Sides of the other. Therefore G H is equal to EH, and AG to BH; and fince A E is equal to EB, and FE is common and at Right Angles, the Bafe A F fhall be f equal to the Bale FB: For the fame Reafon likewise, fhall CF be equal to FD. Again, because AD is equal to C B, and AF to FB, the two Sides FA, AD, will be equal to the two Sides FB, BC, each to each; but the Base DF has been proved equal to the Bafe FC: Therefore the Angle FAD is equal to the Angle FBC: Moreover, AG has been proved equal to BH; but FB alfo is equal to A F. Therefore the two Sides FA, AG, are equal to the two Sides FB, BH; and the Angle FAG is equal to the Angle FBH, as has been demonftrated; wherefore the Bafe G F is † equal to the Bafe FH. Again, becaufe GE has been proved equal to EH, and EF is common, the two Sides GE, EF, are equal to the two Sides HE, EF; but the Bafe HF is equal to the Base FG; therefore the Angle GEF is equal to the Angle HEF, and fo both the Angles GEF, HEF, are Right Angles: Therefore F E makes Right Angles with GH, which is any how drawn thio' E. After the fame Manner we demonstrate that FE is at Right Angles to all Right Lines that are drawn in Def. 3 of the Plane to it; but a Right Line is * at Right Angles to a Plane, when it is at Right Angles to all Right Lines drawn to it in the Plane. Therefore FE is at Right Angles to a Plane drawn thro' the Right Lines

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AB, CD. Wherefore, if to two Right Lines cutting one another, a third ftands at Right Angles in the common Section, it shall be alfo at Right Angles to the Plane drawn thro' the faid Lines; which was to be demon ftrated.

PROPOSITION V.

THEOREM.

If to three Right Lines, touching one another, a third ftands at Right Angles in their common Section, thofe three Right Lines shall be in one and the fame Plane.

LET the Right Line A B stand at Right Angles in the Point of Contact B, to the three Right Lines BC, BD, BE. I say BC, BD, BE, are in one and the fame Plane.

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Def. 3.

For if they are not, let BD, BE, be in one Plane, and BC above it; and let the Plane paffing thro' AB, BC, be produced, and it will make the common * 3 of this. Section, with the other Plane, a ftrait Line, which let be BF. Then three Right Lines AB, BC, BF, are in one Plane drawn thro' AB, BC; and fince AB ftands at Right Angles to BD and BE, it fhall be † at † 4 of this. Right Angles to a Plane drawn thro' BE, DB; and fo A B fhall make Right Angles with all Right Lines touching it that are in the fame Plane; but BF being in the faid Plane, touches it. Wherefore the Angle ABF is a Right Angle, but the Angle ABC (by the Hyp.) is alfo a Right Angle. Therefore the Angle ABF is equal to the Angle ABC, and they are both in the fame Plane, which cannot be; and fo the Right Line BC is not above the Plane paffing thro' BE and BD. Wherefore the three Lines BC, BD, BE, are in one and the fame Plane. Therefore, if to three Right Lines, touching one another, a third ftands at Right Angles in their common Section, thofe three Right Lines fhall be in one and the fame Plane; which was to be demonftrated.

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this.

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PROPOSITIO N VI.

THEOREM.

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

ET two Right Lines AB, CD, be perpendicular to one and the fame Plane. Ifay, 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 DÉ be drawn in the fame Plane at Right Angles; make DE equal to A B, and join BE, AF, AĎ.

Then because AB is at Right Angles to the afore*Def. 3 of 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 Bafe, 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 5 of this. Lines are in one Plane: But BD, DA, are in the +2 of this. fame Planc 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 perpendicular to one and the fame Plane, thofe Right Lines are parallel to one another; which was to be demonftrated.

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PROPOSITION VII,

THEOREM.

If there be two Parallel Lines, and any Points be taken in both of them, the Right Lines joining those 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 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 this: 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 a- 10. 1. bove 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 demonstrated.

PROPOSITION VIII,

THEOREM.

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

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ET AB, CD be two parallel Right Lines, one Se the Fig. of which, as A B is perpendicular to fome of Prop. VI.

Plane. I fay, the other CD is also perpendicular to the

fame Plane.

For let A B, CD, meet the Plane in the Points B, D, 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, and join BE, AE, AD. Then fince AB is perpen

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dicular 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 ABD, 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 Bis equal to DE, and BD is common, the two Sides AB, BD, are equal to the two Sides E D, D B. 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 DÊ, and BE to AD, the two Sides AB, BE, fhall be equal to the two Sides ED, DA, each to each; but the Bafe AE is common. Wherefore the Angle A BE is equal to the Angle EDA; but the Angle A BE is a Right Angle. Therefore EDA is also a Right Angle, and ED is pergendicular to DA; but it is likewife perpendicular to DB: Therefore ED fhall also 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 * 2 of this. AB, BD, are in that Plane; and DC is + in the 17 of this. 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 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 demonstrated.

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