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PROPOSITION XI.

PROBLEM.
To draw a Right Line at Right Angles to a given

Right Line, from a given Point in the same.

ET A B be the given Right Line, and the given

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

Afsume any Point D in AC, and make CE equal * to CD, and upon DE maket the Equilateral 13 of this

ti Triangle FDE, and join FC. I say, the Right Line F C is drawn from the Point C, given in the Right Line A B at Right Angles to A B.

For because DC is equal to CE, and F C is common, the two Lines D C, CF, are each equal to the two Lines EC, CF; and the Base D F is equal to the Base F E. Therefore * the Angle DCF is equal * S of this, to the Angle E CF; and they are adjacent Angles. But when a Right Line, standing upon a Right Line, makes the adjacent Angles equal, each of the equal Angles is a Right Angle; and consequently DCF, | Def. io. FCE, are both Right Angles. Therefore the Right Line FC, &c. which was to be done.

PROPOSITION XII.

PROBLEM.

To draw a Right Line perpendicular, upon a given infinite Right Line, from a Point given out of it.

ET AB be the given infinite Line, and C the

Point given out of it. It is requir’d to draw a Right Line perpendicular upon the given Right Line AB, from the Point C given out of it.

Assume any Point D on the other Side of the Right Line A B, and about the Center C, with the Distance CD describe * a Circle EDG, bisect + EG in H, * Pofl. 3. and join CG, CH, CE. I say there is drawn the + 10 of this,

Per

Perpendicular CH on the given infinite Right Line
AB, from the Point C given out of it.

For because G H is equal to HE, and HC is common, GH and HC are each equal to E H and HC,

and the Base CG is equal to the Base CE. Therefore 18 of this, the Angle CHG is equal | to the Angle CHE; and

they are adjacent Angles. But when a Right Line, standing upon another Right Line, makes the Angles

equal between themselves, each of the equal Angles Def. 10. is a Right one *, and the said standing Right Line is

call'd a Perpendicular to that which it stands on. Therefore CH is drawn perpendicular, upon a given infinite Right Line, from a given Point out of it; which was to be demonstrated.

PROPOSITION XIII.

THEOREM.
When a Right Line, standing upon a Right Line,

makes Angles, these hall be either two Right

Angles, or together equal to two Right Angles. F the

Line CD, make the Angles CBĀ, A BD. I say, the Angles CBA, ABD, are either two Right An

gles, or both together equal to two Right Angles. * Def. E.

For if CB A be equal to ABD, they are * each of + 11 of this them Right Angles : But if not, draw † BE from the

Point B, at Right Angles to CD. Therefore the Angles CBE, EBD, are two Right Angles : And because CBE is equal to both the Angles CBA, ABE,

add the Angle EBD, which is common; and the two I Ax. 2. Angles CBE, EBD, together, are equal to the

three Angles CBA, A BE, EBD, together. Again, because the Angle D BA is equal to the two Angles DBE, EBA, together, add the common Angle ABC, and the two Angles D BA, A B C, are equal to the three Angles D BE, E BA, A B C, together. But it has been prov'd, that the two Angles CBE, EBD, together, are likewise equal to these three Angles : But Things that are equal to one and the same, are * equal between themselves. Therefore likewise the Angles CBE, EBD, together, are equal to the

Angles

* Ax 1.

Angles D BA, A B C, together ; but CBE, EBD, are two Right Angles. Therefore the Angles D BA, ABC, are both together equal to two Right Angles. Wherefore when a Right Line, standing upon another Right Line, makes Angles, these shall be either two Right Angles, or together equal to two Right Angles ; which was to be demonstrated.

PROPOSITION XIV.

THEORE M.

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If to any Right Line, and Point therein, two

Right Lines be drawn from contrary Parts, making the adjacent Angles, both together, equal to two Right Angles, the said two Right Lines

will make but one straight Line. Focom

OR let two Right Lines B C, BD, drawn from

contrary Parts to the Point B, in any Right Line AB, make the adjacent Angles A B C, A BD, both together, equal to two Right Angles. I say, BC, BD, make but one Right Line.

For if BD, CB, do not make one straight Line, let CB and BE make one.

Then, because the Right Line A B stands upon the Right Line CBE, the Angles ABC, A BE, together, will be equal * to two Right Angles. But the Angles, * 13 of this ABC, ABD, together, are also equal to two Right Angles. Now taking away the common Angle ABC, and the remaining Angle A BE is equal to the remaining Angle ABD, the less to the greater, which is impossible. Therefore BE, BC, are not one straight Line. And in the same Manner it is demonstrated, that no other Line but B D is in a straight Line with CB; wherefore CB, BD, shall be in one straight Line. Therefore if to any Right Line, and Point therein two Right Lines be drawn from contrary Parts, making the adjacent Angles, both together, equal to two Right Angles

, the said two Right Lines will make but one straight Line; which was to be demonstrated.

PRO

PROPOSITION XV.

THE ORE M.
If two Right Lines mutually cut each other, the

opposite Angles are equal.

A

ET the two Right Lines A B, CD mutually cut

each other in the Point E. I say, the Angle AEC is equal to the Angle DEB; and the Angle CEB equal to the Angle A ED.

For because the Right Line A E, standing on the

Right Line CD, makes the Angles CEA, AED: * 13 of tbis. These both together shall be equal * to two Right

Angles. Again, because the Right Line D E standing
upon the Right Line AB, makes the Angles AED
DEB: These Angles together are * equal to two Right
Angles. But it has been prov'd, that the Angles CEA,
AĚD, are likewise together equal to two Right An-
gles. Therefore the Angles CEA, AED, are equal

to the Angles A ED, DE B. Take away the common + Ax. 3.

Angle AED, and the Angle remaining CEA, ist
equal to the Angle remaining BED. For the same
Reason, the Angle CEB shall be equal to the Angle
DEA. Therefore if two Right Lines mutually cut
each other, the opposite Angles are equal; which was
to be demonstrated.

Coroll. 1. From hence it is manifest, that two Right

Lines mutually cutting each other, make Angles

at the Section equal to four Right Angles.
Coroll. 2. All the Angles constituted about the same

Point, are equal to four Right Angles.

PRO

PROPOSITION XVI.

THEORE M.
If one side of any Triangle be produced, the out-

ward Angle is greater than either of the inward
opposite Angles.
ET ABC be a Triangle, and one of its Sides

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Angle ACD is greater than either of the inward
Angles CBA, or BAC.

For bisect AC in E *, and join BE, which pro- * 10 of this, duce to F, and make E F equal to B E. Moreover, join F C, and produce A C to G.

Then, because AE is equal to EC, and BE to EF, the two Sides A E, E B, are equal to the two Sides CE, EF, each to each, and the Angle A EB tequal to the Angle FEC; for they are opposite † 15 of this. Angles. Therefore the Base AB, is equal to the 14 of this. Bale FC; and the Triangle A EB, equal to the Triangle F EC; and the remaining Angles of the one, equal to the remaining Angles of the other, each to each, subtending the equal Sides. Wherefore the Angle BAE, is equal to the Angle ECF, but the Angle ECD, is greater than the Angle E CF; therefore the Angle ACD, is greater than the Angle BAE. After the fame manner, if the Right Line BC, be bisected, we demonstrate that the Angle BCG, that is, the Angle A CD, is greater than the Angle ABC. Therefore one Side of any Triangle being produced, the outward Angle is greater than either of the inward opposite Angles; which was to be demonstrated.

PROPOSITION XVII.

THE ORE M.
Two Angles of any Triangle together, howsoever

taken, are less than two Right Angles.

it together, howsoever taken, are less than two Right Angles

For

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