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Join BD, and draw GH bisecting BD at right angles : Then, because the points B, D are in the circumference of each of the circles, the straight line BD falls within each of them (3.2); and therefore the centres of each (3.1. Cor.) must be in the

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straight line GH which bisects BD at right angles; and therefore also GH must pass through the point of contact (3.11): But it does not pass through it, because the points B, D are not in the straight line GH-which is absurd: Therefore one circle cannot touch another on the inside in more points than one.

Nor can two circles touch one another on the outside in more than one point.

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For, if it be possible, let the circle ACK touch the circle ABC in the points A, C, and join AC: Then, because the two points A, C, are in the circumference of the circle ACK, the straight line AC which joins them (3.2) must fall within the circle ACK: And the circle ACK is without the circle ABC; B therefore also the straight line AC is without the circle ABC: But because the two points A, C, are in the circumference of the circle ABC, the straight line AC falls within the circle ABC-which is absurd: Therefore one circle cannot touch another on the outside in more than one point: And it has been shewn, that they cannot touch on the inside in more points than one. Wherefore, One circle &c. Q.E.D.

PROP. XIV. THEOR.

Equal straight lines in a circle are equally distant from the centre; and, conversely, those, which are equally distant from the centre, are equal to one another.

Let the straight lines AB, CD, in the circle ABDC, be equal to one another: they shall be equally distant from the centre.

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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 (3.3); and therefore AF is equal to FB, and AB is double of AF: And, in like manner, it may be shewn that CD is double of CG: But AB is equal to CD (Hyp.); and therefore also AF is equal to CG.

Now, because AE is equal to EC, the square of AE is equal to the square of CE: 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 CG, GE are equal to the square of CE; therefore the squares of AF, FE are equal to the squares of CG, GE: But 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 GE, 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 drawn to them from the centre are equal (3. Def. 4); therefore AB, CD are equally distant from the centre.

Next, let the straight lines AB, CD be equally dis

tant from the centre, that is, let EF be equal to EG: then AB shall be equal to CD.

For, the same construction being made, it may be shewn, as before, that AB is double of AF, and CD of CG, and that the squares of AF, FE are equal to the squares of CG, GE, of which the square of FE is equal to the square of GE, because FE is equal to GE; therefore the remaining square of AF is equal to the remaining square of CG, and the line AF to the line CG: But AB is double of AF, and CD of CG; therefore AB is equal to CD.

Wherefore, Equal straight lines &c. Q.E.D.

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The diameter is the greatest straight line in a circle, and, of all others, that which is nearer to the centre is greater than one more remote: and, conversely, 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 shall be greater than any straight line BC, which is not a diameter, and BC than FG.

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From the centre draw EH, EK, perpendiculars to BC, FG, and join EB, EC, EF: Then, because AE is equal to BE, and ED to EC, therefore the whole AD is equal to the two BE, EC: But BE, EC, are greater than BC; therefore also AD is F greater than BC: Again, because BC is nearer to the centre than FG, therefore EH is less than EK (3. Def. 5): But, as in (3. 14), it may be shewn that BC is double of BH, and FG of FK, and that the squares of EH, HB are equal to the squares EK, KF: But the square of

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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 line BH greater than the line FK, and therefore also BC is greater than FG.

Next, let BC be greater than FG: BC shall be nearer to the centre than FG, that is, the same construction being made, EH shall be less than EK.

For, because BC is greater than FG, therefore also BH is greater than FK: But 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 BII is greater than FK; therefore the square of EH is less than the square of EK, and the line EH than the line EK. Wherefore, The diameter &c.

PROP. XVI.

Q. E.D.

THEOR.

The straight line, drawn at right angles to the diameter of a circle from the extremity of it, falls without the circle: and no straight line can be drawn from the extremity, between that straight line and the circumference, so as not to cut the circle.

Let ABC be a circle, of which the centre 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 not, let it fall, if possible, within the circle, as AC, and draw DC to the point C, where it meets the circumference: Then, because DA is equal to DC, the angle DAC is equal to the angle DCA: But DAC is a right angle; therefore also DCA is a right angle, and the two angles DAC, DCA are therefore

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equal to two right angles-which is impossible (1. 17):

Therefore the straight line drawn from A at right angles to AB does not fall within the circle: And, in like manner, it may be demonstrated that it does not fall upon the circumference: Therefore it must fall without the circle, as AE.

Also, 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 AF be between them; and from the centre D draw DG perpendicular to AF, meeting the circumference in H: Then, because AGD is a right angle, and DAG less than a right angle, therefore DA is greater than DG (1.19): But DA is equal to DH; therefore also DH is greater than DG, the less than the greater-which is absurd: Therefore no straight line can be drawn from the point A between AE and the circumference, so as not to cut the circle. Wherefore, The straight line &c. Q.E. D.

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COR. From this it is manifest, that the straight line, 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 met the circle in two, it would fall within it (3. 2): Also, it is evident that there can be but one straight line, which touches the circle in the same point.

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 the given point A be without the given circle BCD it is required to draw from A a straight line which shall touch the circle BCD.

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