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BF

it is required to draw a straight line from A which shall touch Book III, the circle.

Find a the center E of the circle, and join AE; and from a s. 3. the center E, at the distance EA, describe the circle AFG; from the point D draw b DF at right angles to EA, and join b 11. . EBF, AB; AB touches the circle BCD.

Because E is the center of the circles BCD, AFG, EA is equal to EF: And

A ED to ÉB ; therefore the

D 'two sides AE, EB are equal to the two FE, ED, and they contain the angle at

G CE E common to the two triangles AFB, FED; therefore the base DF is equal to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles": Therefore the C 4. I. angle EBA is equal to the angle EDF: But EDF is a right angle, wherefore EBA is a right angle: And EB is drawn from the center ; but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circled: Therefore d Cor.16.3. AB touches the circle; and it is drawn from the given point A. Which was to be done.

But, if the given point be in the circumference of the circle, as the point D, draw DE to the center E, and DF at right angles to DE; DF touches the circled.

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IF a straight line touches a circle, the straight line drawn

from the center to the point of contact, shall be perpendicular to the line touching the circle.

Let the straight line DE touch the circle ABC in the point C, take the center F, and draw the straight line FC; FC is perpendicular to DE.

For, if it be not, from the point F draw FBG perpendicular to DE ; and because FGC is a right angle, GCF is

b 17. 1. angle; and to the greater angle the greatest lide is opposite : C 19. I.

F2

The efore

an acute

с

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If a straight line touches a circle, and from the point

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

2

a 19. 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 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

A
the point of contact, FC is perpen-
dicular a to DE; therefore FCE is a
right angle : But ACE is also a right

F
angle; therefore the angle FCE is
equal to the angle ACE, the less to B
the greater, which is impossible:
Wherefore F is not the center of the
circle ABC: In the same manner, it

D
may be shewn, that no other point

с E 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.

НЕ
HE 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

a

Le ÁBC be a circle, and BEC an angle at the center, and Book III. 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 center 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

E

a s.r. the angle EBA; therefore the angles EAB, EBA are double of the angle EAB ;

b 32. I. but the angle BEF is equal to the angles B EAB, EBA; therefore also the angle BEF is double of the angle CAB: 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 E the center of the circle be without the angle 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

E 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 center, &c. Q. E. D.

B

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THE

HE angles in the fame segment of a circle are e-Sec N.

qual to one another.

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a

a 20. 3.

Book III. the circumference, viz. BCD, for their base ; therefore the arr

: mgle BFD is double a of the angle BAD: For the same reason,

the angle BFD is double of the angle BED: Therefore the angle BAD is equal to the angle BCD.

But, if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it; these

A E
also are equal to one another : Draw
AF to the center, and produce it to
C, and join CE: Therefore the seg-

В.
ment BADC is greater than a semi-

D
circle; and the angles in it BAC,
BEC are equal, by the first case: For

F
the same reason, because CBED is
greater than a femicircle, the angles
CAD, CED are equal : Therefore
the whole angle BAD is equal to the

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

a

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THE opposite angles of any quadrilateral figure de.

, angles.

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

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

D
is equal 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 ADC is equal to the A

YB
angles CAB, ACB: To each of thefe
equals add the angle ABC; therefore
the angles ABC, CAB, BCA are e-
qual 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,

BAD, DCB may be shewn to be equal to two right angles. Book. III. Therefore, the opposite angles, &c. Q. E. D.

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PON the same straight line, and upon the same fide see N.

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

UPOM

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

D two points A, B, they cannot cut one

a 10. 3. another in any other point ? : One of the segments must thei efore fall within the other; let ACB fall within ADB, and draw the straight line BCD, and join CA, DA: And because the feg.

A

B ment ACB is similar to the segment ADB, and that fimilar segments of circles contain equal an- b 11. def. 3. gles; the angle ACB is equal to the angle ADB, the exterior to the interior, which is impossible. Therefore, there cannot c 16.1. be two similar segmerits of a circle upon the same side of the same line, which do not coincide. Q. E. D.

:

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SI

IMILAR segments of circles upon equal straight See N.

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 segment AEB be

E applied to the segment CFD, so as the point A be on C, and A

в с the straight line AB upon CD, the point B shall coincide with the point D, be. E 4

cause

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