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BOOK III. and join FA, AG: And because F is the centre of the cir
cle ABC, AF is equal
20. 1. But it is also less; which is impossible: Therefore the straight line which joins the points F, G shall not pass otherwise than through the point of contact A, that is, it must pass through it. Therefore, if two circles, &c. Q.E.D.
PROP. XIII. THEOR.
See N. ONE circle cannot touch another in more points than one, whether it touches it on the inside or outside.
For, if it be possible, let the circle EBF touch the circle ABC in more points than one, and first on the inside, in 10. 11. 1. the points B, D; join BD, and drawa GH bisecting BD at right angles: Therefore because the points B, D, are in
Cor. 1. 3.
the circumference of each of the circles, the straight line BD falls within each of them: and their centres are in the straight line GH which bisects BD at right angles: 11. 3. Therefore GH passes through 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 more points than one,
Nor can two cireles a touch one another on the outside Book III.
in more than one point; For, if it be possible, let the cir
cle ACK touch the circle ABC in the
fore one circle cannot touch another on the outside in more than one point. And it has been shown, that they cannot touch on the inside in more points than one. Therefore, one circle, &c. Q. E. D.
PROP. XIV. THEOR.
EQUAL straight lines in a circle are equally distant from the centre; and those which are equally distant from the centre, are equal to one another.
Let the straight lines AB, CD, in the circle ABCD, be equal to one another, they are equally distant from the
Take E the centre of the circle ABDC, and from it draw EF, EG, perpendiculars to AB, CD: Then, because the straight line EF, passing through the centre, cuts the straight line AB, which does not pass through the centre, at right angles, it also bisects it: Wherefore AF is A equal to FB, and AB double of AF. For the same reason CD is double of CG: And AB is equal to CD; there-F fore AF is equal to CG: And because AE is equal to EC, the square of AE is equal to the square of EC: But the squares of AF, FE, are equal to the square of AE, because the 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, FÉ, are
Book III. equal to the squares of CG, GE, of 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 EF is therefore equal to EG: But straight lines in a circle are said to be equally distant from the centre, when the perpendiculars • 4 Def. 3. drawn to them from the centre are equal: Therefore AB, CD must be equally distant from the centre.
Next, if the straight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG; AB is equal to CD: For, the same 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. Therefore equal straight lines, &c. Q. E. D.
PROP. XV. THEOR.
See N. THE diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is always greater than one more remote: and the greater is nearer to the centre than the less.
Let ABCD be a circle, of which the diameter is AD, and the centre E; and let BC be nearer to the centre than FG; AD is greater than any straight line BC, which is not a diameter, and BC greater than FG.
From the centre draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF; and because AE is equal to EB, and ED to EC, AD is
20. 1. equal to EB, EC; but EB, EC are greater than BC: wherefore also AD is greater than BC.
And, because BC is nearer to the centre than FG, EH
b 5 Def. 3.
is less than EK: But, as was demonstrated in the pre- Book III. ceding, BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares of EK, KF, of which the square 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 centre 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 KF: 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.
PROP. XVI. THEOR.
THE 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 centre 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
possible, within the circle, as AC, and draw DC to the point C where it meets the circumference: And
because DA is equal to DC, the B angle DAC is equala to the angle ACD; but DAC is a right angle, therefore ACD is a right angle, and the angles DAC, ACD are
a 5. I.
therefore equal to two right angles; which is impossibleb : ↳ 17. 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; 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 12. 1. 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, 19. 1. DA is greater than DG: But DA
is equal to DH: therefore DH is
COR. From this it is manifest, that the straight line
PROP. XVII. PROB.
To draw a straight 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