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b. 2. 3.

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than one point. for, if it be possible, let the circle ACK touch the Book III. circle ABC in the points A, C, and join AC. therefore because them two points A, C are in the circumference of the circle ACK, the straight line AC

K which joins them shall fall within 6 the circle ACK. and the circle ACK is without the circle ABC, and therefore the

A straight line AC is without this last circle; but because the points A, C, are in the circumference of the circle ABC, the straight line AC must be within the same circle, which is abfurd. therefore one circle cannot touch another on the out -B fide in more than one point. and it has been shewn that they cannot touch on the inside in more points than one. therefore one circle, &c. Q. E. D.

PROP. XIV. THEO R.

EQUAL straight lines in a circle are equally

diftant from the center; and those which are equally distant from the center, are equal to one another.

Let the straight lines AB, CD in the circle ABDC be equal to one another ; they are equally distant from the center.

Take E the center of the circle ABDC, and from it draw FF, EG perpendiculars to AB, CD. then because the straight line EF passing thro' the center cuts the Atraight line AB, which does not pass thro' the center, at right angles, it also bisects it. wherefore AF is equal

a. 3. 3. to FB, and AB double of AF. for the fame reason CD is double of CG. and AB is equal to CD, therefore AF is equal E to CG. and because AE is equal to EC,

E the fquare of AE is equal to the square B of EC. but the squares of AF, FE are equal to the square of AE, because the

b. 47. I. angle AFE is a right angle; and for the like reason the fquares of EG, GC are equal to the square of EC. therefore the squares of AF, FE are equal to the squares of CG,

G

Book III. GE, of which the fquare of AF is equal to the square of CG, be

cause AF is equal to CG ; therefore the remaining square of FE is equal to the remaining square of EG, and the straight line FE is therefore equal to EG. but straight lines in a circle are said to be equally distant from the center, when

the perpendiculars drawn to them from A 6.4. Def. 3 the center are equal . therefore AB,

CD are equally distant from the center.

Next, if the straight lines AB, CD E be equally distant from the center, that

E is, if FE be equal to EG; AB is equal

B to CD. for, the fame construction be ing made, it may, as before, be demonstrated that AB is double of AF and CD double of CG, and that the squares of EF, FA are equal to the squares of EG, GC; of which the square of FE is equal to the square of EG, because FE is equal to EG; therefore the remaining square of AF is equal to the remaining square of CG; and the straight line AF is therefore equal to CG. and AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D.

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PRO P. XV. THEOR, "HE diameter is the greatest straight line in a circle;

and of all others, that which is nearer to the center is always greater than one more remote; and the greater is nearer to the center than the less.

Let ABCD be a circle, of which the diameter is AD, and center E; and let BC be nearer to the center than FG. AD is greater than any straight line BC which is not a

AB diameter, and BC greater than FG.

From the center draw EH, EK per-F pendiculars to BC, FG, and join EB, EC, EF ; and because AE is equal to

IKI

H
EB, and ED to EC, AD is equal to

E
EB, EC. but EB, EC, are greater a than
BC, wherefore also AD is greater than
BC.

And because BC is nearer to the cen D

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ter than FG, EH is less b than EK. but, as was demonstrated in Book III. the preceding, BC is double of BH, and FG double of FK, and w the squares of EH, HB are equal to the squares of EK, KF, of b.s. Dęf. 3. which the fquare of EH is less than the square of EK, because EH is less than EK; therefore the square of BH is greater than the square of FK, and the straight line BH greater than FK; and therefore BC is greater than FG.

Next, let BC be greater than FG; BC is nearer to the center than FG, that is, the same construction being made, EH is less than EK. because BC is greater than FG, BH likewise is greater than FK. and the squares of BH, HE are equal to the squares of FK, KE, of which the square of BH is greater than the square of FK, because BH is greater than FK; therefore the square of EH is less than the square of EK, and the straight line EH less than EK. Wherefore the diameter, &c. Q. E. D.

PRO P. XVI. THEO R. THE 'HE straight line drawn at right angles to the dia-See N.

meter of a circle, from the extremity of it, falls without the circle; and no straight line can be drawn between that straight line and the circumference from the extremity, so as not to cut the circle; or, which is the same thing, no straight line can make so great an acute angle with the diameter at its extremity, or so small an angle with the straight line which is at right angles to it, as not to cut the circle.

Let ABC be a circle the center of which is D, and the diameter AB; the straight line drawn at right angles to AB from its extremity A, shall fall without the circle.

For if it does not, let it fall, if poffible, within the circle as AC, and B

А. draw DC to the point where it

D meets the circumference. and because DA is equal to DC, the angle DAC is equal to the angle ACD;

2. S. 1 but DAC is a right angle, therefore ACD is a right angle, and

E

Book III. the angles DAC, ACD are therefore equal to two right angles ;

w which is impossible 5. therefore the straight line drawn from A at b. 17. s. right angles to BA does not fall within the circle. in the same man

ner it may be demonstrated that it does not fall upon the circumference; therefore it must fall without the circle, as AE.

And between the straight line AE and the circumference no straight line can be drawn from the point A which does not cut the circle. for, if possible, let FA be between them, and from the point D draw DG perpendicular to FA, and let it meet the circumference in H. and because AGD is a right angle, and DAG less than a right angle, DA is great

FU d. 19. 1.

er than DG. but DA is equal to DH;
therefore DH is greater than DG, the

С
less than the greater, which is impof-
sible. therefore no straight line can be
drawn from the point A between AE

B

A and the circumference, which does

D not cut the circle. or, which amounts to the same thing, however great an acute angle a straight line makes with the diameter at the point A, or however small an angle it makes with AE, the circumference passes between that straight line and the perpendicular AE. ' And this is all that is to be understood, ' when in the Greek text and translations from it, the angle of the • femicircle is said to be greater than any acute rectilineal angle, and the remaining angle less than any rectilineal angle.'

COR. From this it is manifest that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle ; and that it touches it only in one point, because if it did meet the circle in two, it would fall within it c. • Also it is evident that there can be but one straight line ! which touches the circle in the same point.'

3.

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To draw a ftraight line from a given point, either

without or in the circumference, which shall touch a given circle.

First, Let A be a given point without the given circle BCD ; it

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

Find the center E of the circle, and join AE; and from the a. 1. 3. center E, at the distance EA describe the circle AFG ; from the point D draw b DF at right angles to EA, and join EBF, AB. b. 11. 1. AB touches the circle BCD.

Because E is the center of
the circles BCD, AFG, EA is
equal to EF, and ED to EB ;

Α.
therefore the two sides AE, EB
are equal to the two FE, ED,
and they contain the angle at G C E
E common to the two triangles
AEB, FED; therefore the base
DF is equal to the bafe AB,
and the triangle FED to the
triangle AEB, and the other
angles to the other angles.

C. 4. I.
therefore the 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 AB 4. Cor.16.3,
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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PRO P. XVIII. THEOR.
IF
F 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 an acute angle; b. 17. S. and to the greater angle the greatest Side is opposite. therefore FCC. 19. 1.

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