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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 * 15. 1.

to two Right Lines CE, EB, and they contain * the 4: 1. equal Angles AED, CEB; the Base A D shall bet

equal to the Base CB, and the Triangle AED equal
to the Triangle CEB; and so likewise is the Angle
DAE equal to the Angle EBC; but the Angle AEG
is * equal to the Angle BEH; therefore AGE,
BEH, are two Triangles, having two Angles of the
one equal to the two Angles of the other, each to each
and one Side A E equal to one Side E B, viz. those

that are at the equal Angles; and so the other Sides 26. .

of the one will be equal to the other Sides of the
other. Therefore GE is equal to EH, and AG to
BH; and fince AE is equal to E B, and FE is com-
mon and at Right Angles, the Base AF shall be +
equal to the Base FB: For the same Reason likewise,
fhall CF be equal to FD. Again, because A D is
equal to CB, and AF to FB, the two Sides FA,
AD, will be equal to the two Sides FB, BC, each

to each; but the Bafe DF has been proved equal to $ 8.5. the Base FC: Therefore the Angle FAD is equal

to the Angle FBC: Moreover, A G has been proved
equal to BH; but F B alfo is equal to AF. There-
fore the two Sides F A, AG, are equal to the two
Sides FB, BH; and the Angle FAG is equal to the
Angle FBH, as has been demonstrated; wherefore
the Bafe GF is equal to the Base FH. Again, be-
cause 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 Base HF is equal to the
Base F G; therefore the Angle GEF is equal to
the Angle HEF, and fo both the Angles GEF,
HEF, are Right Angles: Therefore FE makes Right
Angles with GH, which is any how drawn thro' E.
After the fame manner we demonstrate that FE is

at Right Angles to all Right Lines that are drawn in * Dec 3.of the Plane to it; but a Right Line is * at Right Angles tbiso

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

AB,

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AB, CD. Wherefore, if to two Right Lines cutting one another, a third stands at Right Angles in the common Section, it shall be also at Right Angles to the Plane drawn thro' the said Lines; which was to be demonstrated.

PROPOSITION V.

THE OR E-M.
If to three Right Lines, touching one another, a

third stands at Right Angles in their common
Sedion, those three Right Lines shall be in one
and the same Plane.
ET the Right Line A B stand at Right Angles

in the Point of Contact B, to the three Right Lines B C, BD, BE. I say BC, BD, BE, are in one and the fame Plane.

For if they are not, let BD, BE, be in one Plane; and BC above it; and let the Plane passing thro' AB, BC, be produced, and it will * make the common * 3 of this, Section, with the other Plane, a strait Line, which det be BF. Then three Right Lines AB, BC, BF, are in one Plane drawn thro' AB, BC; and fince AB stands at Right Angles to BD and BE, it shall be + at Right Angles to a Plane drawn thro' BE; + 4 of this DB; and to AB shall make | Right Angles with I Def. 3. all Right Lines touching it that are in the same 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 also a Right Angle. Therefore the Angle ABF is equal to the Angle ABC, and they are both in the same Plane, which cannot be; and so the Right Line B C is not above the Plane paffing thro' BÈ and BD. Wherefore the three Lines BC, BD, BE, are in one and the fame Plane.

Therefore, if to thru Right Lines, touching one another, a third stands at Right Angles in their common Section, those three Right Lines shall be in one and the fame Plane'; which was to be demonstrated.

PROPOSITION VI.

THEOREM.
If two Right Lines be perpendicular to one and the

Same Plane, those Right Lines are parallel to
one anotber.

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ET two Right Lines AB, CD, be perpendicu

this.

parallel to CD.

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

Then because AB is at Right Angles to the aforeDef. 3. of said Plane, it shall 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 Reason likewise, 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, shall be equal to the two Sides ED, DB;

but they contain Right Angles. Therefore the Base + 4. 1. AD ist equal to the Base 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 18. 1.

is I equal to the Angle EDA; but A BE is a Right Angle. Therefore EDA is also a Right Angle;

and fo ED is perpendicular to DÄ; but it is also

perpendicular to BD and DC. Therefore E D is at Right Angles in the Point of Contact to three Right Lines

BD, DA, DC. Wherefore these three last Right 5 of this. Lines are * in one Plane : But BD, DA, are in the + 2 of Ibis. fame Plane as AB is; for every Triangle is t in the

same Plane. Therefore it is neceffary that AB, BD,

DC, be in one Plane ; but both the Angles ABD, I 28. 1. BDC, are Right Angles. Wherefore AB is paral

lel to CD. Therefore, if two Right Lines be perpendicular to one and the fame Plane, those Right Lines are parallel to one another; which was to be demonstrated.

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PROPOSITION VII,

THE ORE M. If there be two Parallel Lines, and any Point be taken in both of them, the Right Lines joining those Points shall be in the same Planes as the Parallels are.

which are taken any Points E, F. I say, a Right Line joining the Points É, F, are in the fame Plane as the Parallels are,

For if it be not, let it be elevated above the same, if possible, 'aş" 'EGF; thro' which let some Plane bé drawn, whöfe Section, with the Plane in which the two Right Lines EGF, EF, will include a Space, which is f absurd. Therefore a Right' Line drawn † Axiom from the Point E to the Point F, is not elevated 10. 1. above the Plane, and consequently it must be in that passing thro' the Parallels AB, C D. Wherefore, if there be two parallel Lines, and any Points be taken in both of them, the Right Line joining these Points fhall be in the fame Plane as the Parallels are ; which was to be demonstrated,

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PROPOSITION VIII.

T # É O R E M.
If there be two parallel Right Lines, one of which

is perpendicular to some Plane, then mall the

other be perpendicular to the same Plane. LET

ET AB, CD, be two parallel Right Lines, one See tbe Fig.

of which, as AB is perpendicular to some of Prop. VI. Plane. I say, the other CD is also perpendicular to the same Plane.

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

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

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* Def. 3.

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 same Plane; therefore each of the Angles ABD, ABE, is a Right Angle. And since the Right Line BD falls on

the Right Lines AB, CD, the Angles A BD, CDB, +29. 1. fhall be + equal to two Right Angles. Therefore the

Angle CDB is also a Right Angle, and so CD iş perpendicular to DB: And since AB 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 14. 1. is a Right Angle. Therefore the Base AD is I equal to

the Base BE. Again, since AB is equal to DE, and BE to AD, the two Sides AB, BE, shall be equal to the two Sides ED, DA, each to cach; but the

Base A E is common. Wherefore the Angle ABE * 8.1. equal to the Angle E DA;, but the Angle ABE

is a Right Angle. Therefore EDA is also a Right An

gle, and ED is perpendicular to DA; but it is like... wise perpendicular to DB: Therefore ED shall also + 4. f this. be f perpendicular to the Plane passing thro' BD, DA, Def . 3. and likewise, shall be I at Right Angles to all Right

Lines, drawn in the said Plane that touch it. But

DC is in the Plane passing thro' BD, DA, because * 2 of this. AB, BD, are * in that Plane; and DC ist in the to this. fame Plane that AB and BD are in. Wherefore

ED is at Right Angles to DC, and so CD is at Right Angles to DE, as also to D B. Therefore CD stands 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 passing thro' DE, D B; which was to be demonstrated.

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