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Book III. join FA, AG: And because F is the center of the circle ABC,
See N. :
PRO P. XIII. THEOR.
one, whether it touches it on the inside or outfide.
· For, if it be possible, let the circle EBF touch the circle
ABC in more points'than one, and first on the inside, in the 1 10.11.1. points B, D; join BD, and draw.GH bisecting BD at right angles.: Therefore, because the points B, D are in the circumfe
b 2. 3.
G rence of each of the circles, the ftraight line BD falls within
o each of them: And their centers are in the straight line-Gli c Cor. 1. 3. which bifects BD at right angles; therefore GH pafles through d 11. 3. the point of contactd, but it does not pass through it, because the
points B, D are without the straight line GH, which is absurd:
Therefore one circle cannot touch another on the inside in
b 2. 3.
more than one point : For, if it be possible, let the circle ACK Book IIT. touch the circle ABC in the points A, C, and join AC: There i fore, because the two points A, C are in the circumference of the circle ACK,
K the straight line AC which joins them shall fall within the circle ACK : And the circle ACK is without the circle ABC; and therefore the 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 withirr b 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.
QUAL straight lines in a circle are equally distant
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 EF, EG perpendiculars to AB, CD: Then, because the straight line Er passing through the center cuts the straight line AB, which does not pass thro' the center, at right angles, it also bisects a it: Wherefore
a 3. 3. AF is equal to FB, and AB double of AF. For the same reason, CD is dou
А. ble of CG: And AB is equal to CD; therefore AF is equal to CG! And because AE is equal to EC, the square F of AE is equal to the square of EC: But the squares of 'AF, FE are equal
147. I. o to the square of AÉ, because the B angle AFE is a right angle; and, for the like reason, the squares of EG, GC are equal to the square of EC: Therefore the squares of AF, FE are equal to the squares of CG, GE, of
Book III. which the square of AF is equal to the square of CG, because
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 64. Def. 3. drawn to them from the center are equal : Therefore AB, CD
are equally distant from the center.
Next, if the straight lines AB, CD be equally distant from the center, that is, if FE be equal to EG; AB is equal to CD: For, the fame construction being 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. There. fore equal straight lines, &c. Q. E. D.
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
EH is less ó than EK: But, as was demonstrated in the pre
Book IIT. ceeding, BC is double of BH, and FG double of FK, and the m squares of EH, HB are equal to the squares of EK, KF, of bs. Def. 3. which the square of EH is less than the fquare of EK, because EH is less than EK; therefure 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, Eki 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 TK, 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.
"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 itraight 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 fo 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 P, 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 poflint, within the circle as AC, and d DC to the point C where it meets the circumference: And because DA is equal to DC, the B angle DAC is equal to the angle
a AČID; but DAC is a right angle, there Kore ACD is a right angle, and the angles DAC, ACD are therefore equal to two right angles, which is impossible 6: 117, 1,
A a 5.1.
C 12. 1.
Book III. Therefore the straight line drawn from A at right angles to BA
does not fall within the circle: In the same manner, it may be demonstrated that it does not fall upon the circumference; there. fore 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 d 19. I. angle, and DAG less than a right angle: DA is greater than DĠ : But DA is equal to DH;
F therefore DH is greater than DG,
E the less than the greater, which is
D 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 pafies 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 semicircle is said to be greater than any acute rectilineal
angle, and the remaining angle less than any rectilineal an. 'gle.'
Cor. From this it is manisest that the straight line which is drawn at right angles to the diameter of a circle from the ex. tremity of ii, 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 e. Also it is evident that there can be but one • straight line which touches the circle in the same point.'
e 2. 3.
To draw a straight line from a given point, ither
without or in the circumference, which shall to lich a given circle.
First, Let A be a given point without the given circle ICD.