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IF a straight line touch'a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle shall be in chat line.

* 18. 3.

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

For, if not, let F be the centre, if possible, and join CF: because DE touches the circle

A
ABC, and FC is drawn from the
centre to the point of contact, FC
is perpendicular a to DE; there.
fore FCE is a right angle: but

F
ACE is also a right angle; there. B
fore the angle FCE is equal to
the angle' ACE, the less to the
greater, which is impossible:
wherefore F is not the centre of
the circle ABC: in the same D

С

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

PROP. XX. THEOR.

See Note.

THE angle at the centre of a circle is double of the

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

Let ABC be a circle, and BEC an angle at the centre, and Book 111. BAC an'angle at the circumference, which have the same circumference BC for their base; the angle

A
BEC is double of the angle BAC. /

First, Let E the centre of the circle be
within the angle BAC, and join AE, and
produce it to F: because EA is equal to
EB, the angle EAB is equal to the

E

a 5. 1. angle EBA;

therefore the angles EAB, FBA are double of the angle EAB; but the angle BEF is, equal to the angles

b 32... EAB, EBA; therefore also the angle BEF

с
is double of the angle EAB : for the same F
reason, the angle FEC is double of the angle EAC: therefore
the whole angle BEC is double of the whole angle BAC.
Again, Let 'E the centre of the

A
circle be without the angle BDC, and
join DE and produce it to G. In

D
may be demonstrated, as in the first
case, that the angle GEC is double

E
of the angle GDC, and that GEB a
part of the first is double of GDB a
part of the other; therefore the re-G
maining angle BEC is double of the
remaining angle BDC. Therefore
the angle at the centre, &c. Q. E. D. B

a

PROP. XXI. THEOR. THE angles in the same segment of a circle are See Note: equal to one another.

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Book 111. the circumference, viz. BCD, for their base; therefore the an

no gle BFD) is double a of the angle BAD: for the same reason, a 20.3. the angle BFD is double of the angle BED: therefore the an

gle 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

А. E
also are equal to one another : draw
AF to the centre, and produce it to
C, and join ÇE: therefore the seg-B

D
ment BADC is greater than a semi-
circle; and the angles in it, BAC,

F
BEC are equal, by the first case : for
the same reason, because CBED is
greater than a semicircle, the angles
CAD, CED are equal : therefore the

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

40th L.

PROP. XXII. THEOR.

THE opposite angles of any quadrilateral figure described in a circle, are together equal to two right angles.

a 52. 1.

b 21. 3.

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

Join AC, BD;, and because the thrce angles of every triangle are equal a to two right angles, the three angles of the triangle ÇAB, viz. the angles CAB, ABC, BCA are equal to two right angles: but the angle CAB

D
is equal b to the angle CDB, because
they are in the same segment BADC,
and the angle ACB is equal to the
angle ADB, because they are in the
same segment ADCB: therefore the
whole angle ABC is equal to the A

B
angles CAB, ACB: to each of these
equals add the angle ABC; therefore
the angles ABC, CAB, BCA are
cqua) 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 shown to be equal to two right angles. Book III.
Therefore the opposite angles, &c. Q. E. D.

1

PROP. XXIII. TIIEOR.

UPON the same straight line, and upon the same See Note: side of it, there cannot be two similar segments of circles, not coinciding with one another.

:

a 10. 3.

If it be possible, let the two similar segments of circles, viz.
ACB, ABD be upon the same side of the same straight line
AB, not coinciding with one another : then, because the cir-
'cle ACB cuts the circle ADB in the two
points A, B, they cannot cut one another

D
in any other pointa : one of the segments
must therefore fall within the other; let
ACB fall withinADB, and draw the straight
line BCD, and join CA, DA: and because
the segment ACB is similar to the segment A

B
ADB, and that similar segments of circles contain equal an-b 11. desf.
gles; the angle ACB is equal to the angle ADB, the exterior
to the interior, which is impossible c. Therefore, there can-c 16 1.
not be two similar segments of a circle upon the same side of
the same line, which do not coincide. Q. E. D.

с

PROP. XXIV: THEOR.

dies

SIMILAR segments of circles upon equal straight See Note. lines, are equal to one another.

Let AEB, CFD be similar segments of circles upon the
equal straight lines AB, CD: the segment AEB is equal to
the segment CFD.
For, if the seg.

E

F
ment AEB be ap-
plied to the seg.
ment 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),

.

Book-311. because AB is equal to CD: therefore' the straight line AB

coinciding with CD, the segment AEB must a coincide with a 23. 3. the segment CFD, and therefore is equal to it. Wherefore,

similar segments, &c. Q. E. D.

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PROP. XXV. PROB,

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

\ * 10, 1.

b 11. 1.

c 6.1.

с

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

Bišecta AC in D, and from the point D drawb DB at right 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, DC, are all equal ; D is the centre of the circled: from the centre D, at the distance of any of the three DA, DB, DC, describe a circle ; this shall pass through the other points; and the circle of which ABC is a segment is described: and because the centre D is in AC, the segment ABC

d 9. 3.

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a

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e 23. 1.

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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 e the angle BAE equal to the angle ABD, and produce BD, if necessary; 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 be cause AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two ČD, DE, each to each; and the angle ADE is equal to the angle CDE, for each of them is a right angle; therefore the base AE is equals to the base EC: but AE was shown to be equal to EB, wherefore also BE is equal to EC: and the three straight lines

14.1.

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