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PROPOSITION IX.

PROBLEM.

To cut a given Right-lin'd Angle into two equal

Parts.

LET BAC be a given Right-lin'd Angle, which

is required to be cut into two equal Parts.

Afsume any Point D in the Right Line A B, and + 3 of tbis, cut off A E from the Line A C equal to A D ti join 11 of this. D E, and thereon make | the equilateral Triangle

DE F, and join AF. I lay, the Angle BAC is cut
into two equal Parts by the Line A F.

For, becaule A D is equal to A E, and A F is com-
mon, the two Sides DA, AF, are equal to the

two Sides A E, A F, and the Base D F is equal to the 1 8 of tbis. Base E F; therefore I the Angle DAF is equal to

the Angle EAF. Wherefore, a given Right-lin'd
Angle is cut into two equal Parts ; which was to be done.

PROPOSITION X. .

2

PROBLEM.
To cut a given finite Right Line into two equal

Parts.

LET A B be a given finite Right Line, required

to be cut into two equal Parts.
* 1 of ibis. Upon it make * an Equilateral Triangle A B C, and
tooftbis

. bisect + the Angle ACB by the Right Line CD. I
say the Right Line A B is bifected in the Point D.

For, because AC is equal to CB, and C D is com-
mon, the Right Lines A C, CD, are equal to the

two Right Lines BC, CD, and the Angle ACD 4 of tbis. equal to the Angle BCD; therefore I the Base AD

is equal to the Bale D B. And so, the Right Line
A B is biseated in the Point D; which was to be
done.

PRO

PROPOSITION XI.

1

PROBLEM
To draw a Right Line at Right Angles to a given
Right Line, from a given Point in the same.

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

Afsume any Point D in A C, and make CE equal * to CD; and upon D E make + the Equilateral • 3 of tbis. Triangle FDE, and join FC. I say, the Right 1 of tbis, 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 equal to the two Lines E C, CF; and the Base D F is equal to the Base F E. Therefore * the Angle DCF is equal * 8 of tbiro to the Angle ECF; 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 I a Right Angle; and consequently DCF, t 10 of ibis. FC E, are both Right Angles. Therefore, the Right Line F C is drawn from the Point C at Right Angles to A B; which was to be done.

PR POPSITION XII.

PROBLEM.
To draw a Right Line perpendicular, upon a given

infinite Right Line, from a Point given out of it. LET A B 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 Centre C, with the Diftance CD, describe * a Circle EDG; bisect + E G in H, . Poft. 3. and join CG, CH, CE. I say, there is drawn the t 10 ofibis. *

Per

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

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

and the Bafe CG is equal to the Bale CE. Therefore $ 8 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

called a Perpendicular to that which it Itands on. Therefore, C H is drawn perpendicular upon a given infinite Right Line, from a given Point out of it; which was to be done.

PROPOSITION XIII.

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

makes Angles, these shall be either two Right
Angles, or together equal to two Right Angles.
O R let a Right Line A B, standing upon the Right

Line CD, make the Angles CBA, ABD. I say, the Angles CBA, A BD, are either two Right Angles, or both together equal to two Right Angles.

For if C B A be equal to A B D, they are * each of + 11 of this, them Right Angles : But if not, draw + B E from tha

Point B, at Right Angles to CD. Therefore the Angles Ç BE, EBD, are two Right Angles : And because CBE is equal to both the Angles CBA, ABE, add the Angle E BD, which is common; and the two Angles CBE, EBD, together, are f equal to the three Angles CBA, ABE, EBD, together, Again, because the Angle DBA is equal to the two Angles D BE, E B A, together, add the common Angle A BC, and the two Angles D BA, ABC, are equal to the three Angles D B E, EBA, A B C, together. But it has been proved, 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. Therefire likewise the Angle C BE, EBD, together, are equal

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to the Angle B DA, ABC, together; but CBE, EBD, are two Right Angles. Therefore the Angles DBA, A B C, 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.

THEOREM.
If to any Right Line, and Point therein, two

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

Lines will make but one strait Line.
FOR

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

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

For if BD, CB, do not make one strait Line, let CB and B E make one.

Then, because the Right Line AB stands upon the Right Line C BE, the Angle ABC, AB E, together, will be equal * to two Right Angles. But the Angles • 13 of ibis. ABC, ABD, together are also equal to two Right Angles. Now taking away the common Angle A B C, the remaining Angle A B E is equal to the remaining Angle A BD, the less to the greater, which is impossible. Therefore BE, BC, are not one ftrait Line. And in the fame manner it is demonstrated, that no other Line but B D is in a strait Line with CB; wherefore CB, BD, fhall be in one strait, 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 Arait Line; which was to be demonstrated.

PRO

PROPOSITION XV.

THEOREM.

If two Right Lines mutually cut each other, the

opposite Angles are equal.

LE T the two Right Lines A B C D, mutually cut

each other in the Point E. I say, the Angle Ą EC is equal to the Angle D E B; and the Angle CEB equal to the Angle A E D.

For because the Right Line A E, standing on the

Right Line CD, makes the Angles CEA, AED; 13 of this, these both together fall be equal * to two Right Angles. Again,

because the Right Line D Е, standing upon the Right Line A B, makes the Angles A ED, DEB; these Angles together are * equal to two Right Angles. But it has been proved that the Angles CEA, A É D, are likewise together equal to two Right Angles. Therefore the Angles CEA, A E D, are equal to the Angles A ED, DEB. Take away the com

mon Angle A ED, and the Angle remaining CE A is + Ax. 20

+ equal to the Angle remaining BED. For the same Reason, the Angle CE B shall be equal to the Angle DEA. Theretore, if two Right Lines mutually cut each other, the opposite Angles are equal; which was to be deinonstrated.

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

Lines, inutually cutting each other, make Angles

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

Point, are equal to four Right Angles.

PRO

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