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Book III.

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 ABDC, be equal to one another; they are equally distant from the centre.

Take E the centre of the circle ABDC, and from it draw EF, EG, perpendiculars to AB, CD; join AE and EC. 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 AFA
is equal to FB, and AB double
of AF. For the same reason, CD
is double of CG: But AB is
equal to CD; therefore AF is
equal to CG: And because AE
is equal to EC, the square of AE

b

F

E

C a 3. 3.

G

B D

is equal to the square of EC: Now the squares of AF, FE are equal to the square of AE, because the angle b 47. 1. 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 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 drawn to them from the centre are equal: Therefore e 3. Def. 3. AB, CD are 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

G

Book III. 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: But AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D.

a 20. 1.

PROP. XV. THEOR.

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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 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 cen-
tre E; and let BC be nearer to F
the centre than FG; AD is greater
than any straight line BC which is
not a diameter, and BC greater K
than FG.

From the centre draw EH, EK
perpendiculars to BC, FG, and
join EB, EC, EF; and because AE

a

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is equal to EB, and ED to EC, AD is equal to EB, EC: But EB, EC are greater than BC; wherefore, also AD is greater than BC.

b

And because BC is nearer to the centre than FG, 4. Def. 3. EH is less than EK: but as was demonstrated in the preceding, 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 Eй 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 Book III. FG, BH likewise is greater than KF; 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 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 diameter 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 of the diameter, so as not to cut the circle.

Let ABC be a circle, the centre of which is D, and the diameter AB; and let AE be drawn from A perpendicular to AB, AE shall fall without the circle.

IE

á 32. 1. Fb. 19. 1.

D

A

In AE take any point F, join DF, and let DF meet the circle in C. Because DAF is a right angle, it is greater than the angle AFDa; but the greater angle of any triangle is subtended by the greater side", therefore DF is greater than DA; now DA is equal to DC, B therefore DF is greater than DC, and the point F is therefore without the circle. And F is any point whatever in the line

AE, therefore AE falls without the circle.

Again, between the straight line AE and the circumference, no straight line can be drawn from the point A, which does not cut the circle. Let AG be drawn, in

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Book III. the angle DAE; from D draw
DH at right angles to AG;
and because the angle DHA is
a right angle, and the angle
DAH less than a right angle,
the side DH of the triangle
DAH, is less than the side B
DA. The point H, there-
fore, is within the circle, and
therefore the straight line AG
cuts the circle."

e 2. 3.

à 1. 3,

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

b

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

Because E is the centre of
the circles BCD, AFG, EA
is equal to EF, and ED to
EB; therefore the two sides G/
AE, EB are equal to the two
FE, ED, and they contain
the angle at E common to the
two triangles AEB, FED;
therefore the base DF is equal
to the base AB, and the trian-
gle FED to the triangle AEB,

A

AVA

D

E

B

OF GEOMETRY.

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ו'ד י

101

c 4. 1.

and the other angles to the other angles : Therefore Book III. 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 centre: but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circled Therefore AB touches the circle; and is a Cor. 16. 3. 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 centre E, and DF at right angles to DE; DF touches the circle.

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If a straight line touch a circle, the straight line drawn from the centre to the point of contact is perpendicular to the line touching the circle.

Let the straight line DE touch the circle ABC in the point C; take the centre 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 must be an acute angle; and to the greater angle the greater side is opposite: Therefore FC is greater than FG; but FC is equal to FB; therefore FB is greater than FG, the less than the greater, which is impossible; wherefore FG is not perpendicular to DE: In the same manner it may be shewn, that no other line but

F

B

C G

E

FC can be perpendicular to DE; FC is therefore per

pendicular to DE.

Therefore, if a straight line, &c.

b 17. 1.

c 19. 1.

Q. E. D.

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