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Book III. is greater than FG; but FC is e

A
qual to FB; therefore FB is greater
than FG, the less than the greater,
which is impossible. wherefore FG
is not perpendicular to DE, in

F
the same manner it may be shewn,
that no other is perpendicular to
it besides FC, that is, FC is per-
pendicular to DE. Therefore if a

D C GE straight line, &c. Q. E. D.

PROP. XIX. THEOR.
IF
F a straight line touches a circle, and from the point of

contact a straight line be drawn at right angles to the touching line, the center of the circle shall be in that line.

Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE; the center of the circle is in CA.

For if not, let F be the center, if possible, and join CF. Because DE touches the circle ABC, and FC is drawn from the center to the point

A 2. 18. 3. of contact, FC is perpendicular a to

DE; therefore FCE is a right angle. .
but ACE is also a right angle; there-

F
fore the angle FCE is equal to the

B
angle ACE, the less to the greater,
which is impossible. wherefore F is
nor the center of the circle ABC. in

D C E
the fame manner. it may be thewn
that no other point which is not in CA, is the center; that is, the
center is in CA. Therefore if a straight line, &c. Q. E. D.

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See N. THE angle at the center of a circle is double of the

angle at the circumference, upon the same base, that is, upon the same part of the circumference,

Let

2. 5. I.

Let ABC be a circle, and BEC an angle at the center, and BAC Book III. an angle at the circumference, which have the same circumference BC for their base; the angle BEC is double

A. of the angle BAC.

First, Let E the center of the circle be within the angle BAC, and join AE, and produce it to F. Because E A is equal to

E
EB, the angle EAB is equal a to the angle
EBA; therefore the angles E AB, EBA
are double of the angle EAB; but the B

с angle BEF is equal b to the angles EAB,

T

b. 32. 1. EBA; therefore also the angle BEF is double of the angle EAB. for the fame reason, the angle FEC is double of the angle EAC. therefore the whole angle BEC is double of the whole angle BAC.

Again, Let BDC be inflected to the circumference, so that E the conter of the circle be without the angle

D BDC, and join DE and produce it to G. It may be demonstrated, as in the first case, that the angle GEC is double of the angle GDC, and that GEB a

G part of the first is double of GDB a part of the other ; therefore the re

B В С maining angle BEC is double of the remaining angle BDC. Therefore the angle at the center,&c.Q.E.D.

PROP. XXI. THEOR.

E

THE angles in the same segment of a circle are equal see N.

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to one another. Let ABCD be a circle, and BAD, BED angles in the fame segment BAED; the angles BAD, BED are equal to one another.

Take F the center of the circle ABCD, and, first, let the fegment BAED be greater than a semicircle, B and join BF, FD, and because the angle BFD is at the center, and the angle BAD at the circumference, and

D

that

Book III. that they have the same part of the circumference, viz. BCD for mtheir base, therefore the angle BFD is double of the angle BAD. 2. 20. 3. for the same reason, the angle BFD is double of the angle BED.

therefore the angle BAD is equal to the angle BED.

But if the segment BAED be not greater than a semicircle, let
BAD, BED be angles in it; these
also are equal to one another. draw

A E
AF to the center, and produce it to
C, and join CE. therefore the segment B
BADC is greater than a semicircle ;
and the angles in it BAC, BEC are e-

F
qual, by the first case. for the same
reason, the angles CAD, CED are
equal. therefore the whole angle BAD

С
is equal to the whole angle BED.
Wherefore the angles in the same segment, &c. Q. E. D.

D

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THE opposite angles of any quadrilateral figure de

scribed in a circle, are together equal to two right angles.

Let ABCD be a quadrilateral figure in the circle ABCD; aný two of its opposite angles are together equal to two right angles.

Join AC, BD; and because the three angles of every triangle are 2. 32. 1. equal to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA are equal

D
to two right angles. but the angle
b.21. 3. CAB is equal o to the angle CDB, be-

cause they are in the fame segment
BADC; and the angle ACB is equal
to the angle ADB, because they are in

A

B the same segment ADCB. therefore the whole angle ADC is equal to the angles CAB, ACB. to each of these equals add the angle ABC, therefore the angles ABC, CAB, BCA are equal to the angles ABC, ADC, but ABC, CAB, BCA are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles. in the same manner the angles BAD, DCB may be

Mewn

thewn to be equal to two right angles. Therefore the opposite Book III. angles, &c. Q. E. D.

PROP. XXIII. THEOR. UPON the fame straight line, and upon the same fide see N

of it, there cannot be two fimilar segments of circles, not coinciding with one another.

2. 10. 3.

If it be possible, let the two similar segments of circles, viz. ACB, ADB be upon the same side of the same straight line AB, not coinciding with one another. then because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point a. one of the segments must therefore fall within the other ; let ACB fall within ADB, and draw the straight line BCD, and join CA, DA. and because the seg. A

B ment ACB is similar to the segment ADB, and that similar segments of circles contain equal angles ; the.b.11.Def.3. angle ACB is equal to the angle ADB, the exterior to the interior, which is impossible. Therefore there cannot be two similar seg. c. 16. s. ments of a circle upon the same side of the same line, which do not coincide. Q. E. D.

PROP. XXIV. THEOR.

SImilar fegments of circles upon equal straight lines, See N.

are equal to one another.

Let AEB, CFD be fimilar segments of circles upon the equal Atraight lines AB, CD; the segment AEB is equal to the segment CFD.

For if the seg. ment AEB be apo

E plied to the fegment CFD, so as the point A be on A

B C

D C,and the Straight line AB upon CD, the point B shall coincide with the point D, be

cause

Book III. cause AB is equal to CD. therefore the straight line AB coinciding

with CD, the segment AEB must coincide with the segment 2. 23. 3. CFD, and therefore is equal to it. Wherefore similar segments, &c.

Q. E. D.

PROP. XXV. PROB.

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Segment of a circle being given, to describe the circle of which it is the segment.

a. 1o. I.

c. 6. I.

Let ABC be the given segment of a circle'; it is required to defcribe the circle of which it is the segment.

Bisect a AC in D, and from the point D drawb DB at right b. 11. . angles to AC, and join AB. First, let the angles ABD, BAD be”

equal to one another ; then the straight line BD is equal < to DA,

and therefore to DC, and because the three straight lines DA, DB, d. 9. 3. DC are all equal, D is the center of the circled from the center

D, at the distance of any of the three DA, DB, DC describe a circle; this shall pass thro’ the other points; and the circle of which ABC is a fegment is described. and because the center D is in AC,

3 B

B

B.
E

D

A
A

A
D

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D

the segment ABC is a semicircle. but if the angles ABD, BAD are

not equal to one another, at the point A in the straight line AB make c. 23. 1. the angle BAE equal to the angle ABD, and produce BD to E, and

join EC. and because the angle ABE is equal to the angle BAE; the straight line BE is equal to EA. and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each; and the

angle ADE is equal to the angle CDE, for each of them is a right f. 4. 1. angle; therefore the base AE is equal to the base EC. but AE

was shewn to be equal to EB, wherefore also BE is equal to EC ; and the three straight lines AE, EB, EC are therefore equal to one

another;

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