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Book III. square of CG; therefore the straight line AF is equal to CG.

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

PROP. XV. THEOR.

THE diameter is the greatest straight line in a circle ; and of all other lines 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.

Prof. Kali

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Let ABCD be a circle, of which the diameter is AD, and the centre

A

B В
E ; and let BC be nearer to the
centre than FG; AD is greater than F
any straight line BC, which is not a
diameter, and BC greater than FG.
From the centre draw EH, EK

E
perpendicular to BC, FG, and join
EB, EC, EF. Because AE is
equal to EB, and ED to EC, AD

is equal to EB, EC. But EB, EC a 30. 1. are greatera than BC; wherefore also AD is greater than BC. b 4. Def. 3.

Because BC is nearer to the centre than FG, EH is lessb

than EK. BC is double of BH', and FG double of FK; c 3. 3.

and the squares of EH, HB are equal to the squares of EK, KF, because they are equal to the squares of the radii EB, EF. But 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; therefore the straight line BH is greater than FK; therefore BC is greater than FG.

Next, let BC be greater than FG; BC is nearer to the centre than FG; that is, EH is less than EK. Because BC is greater than FG, BH is greater than KF. But the squares of BH, HE are equal to the squares of FK, KE, and the square of BH is greater than the square of FK; therefore the square of EŇ is less than the square of EK; therefore the straight line EH is less than EK. Wherefore, the diameter, &c. Q. E. D.

PROP. XVI. THEOR.

Book III.

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 will fall without the circle.

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 AFD«; therefore DF is B

a 32. 1. D

ΙΑ greater than DAb or DC; there

b 19.1. fore the point F is 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 the angle DAE; from D draw DH at right angles to AG. 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 DAb; therefore the point H is within the circle, and therefore B

A

D the straight line AG cuts the circle. Therefore, the straight line, &c. Q. E. D.

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Book III.

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 points, 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.

C 2. 3.

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. a 1.3. Finda the centre E of the circle, and join AE ; from the

centre E, at the distance EA, describe the circle AFG; from b 11. 1. the point D drawb DF at right angles to EA, and join EBF

and AB. AB touches the circle BCD.

Because E is the centre of the
circles BCD, AFG, EA is equal

D
to EF, and ED to EB; there.
fore the two sides AE, EB are
equal to the two FE, ED, and
they contain the angle at E com-

14

a mon to the two triangles AEB,

FED; therefore the angle EBA C 4. 1. is equal to the angle EDFc. But

EDF is a right angle, wherefore

EBA is a right angle; and EB is drawn from the centre; d Cor.16.3. therefore AB touches the circled, 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 centre E, and DF at right angles to DE; DF touches the circled.

PROP. XVIII. THEOR.

Book III.

IF a straight line touch a circle, the straight line drawn from the centre to the point of contact is per: pendicular to the line touching the circle.

a 17. 1.

b 19. 1.

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

A is a right angle, GCF must bea an acute angle ; therefore FC is

greater than FGb. But FC is equal to FB; therefore FB

F is greater than FG, the less than the greater, which is impossible; wherefore FG is not

B perpendicular to DE. In the same manner it

may

be shown that no other line but FC can

D

C:G E be perpendicular to DE; therefore FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D.

PROP. XIX. THEOR.

IF a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the centre of the circle is in that line.

Book III. Let the straight line DE touch the circle ABC in C, and

from C let CA be drawn at right angles to DE; the centre of
the circle is in CA.

For, if not, let F be the centre, if possible, and join CF.
Because DE touches the circle
ABC, and FC is drawn from

the centre to the point of cona 18. 3. tact, FC is perpendiculara to

DE; therefore FCE is a right
angle. But ACE is also a

F
right angle; therefore the an B
gle FCE is equal to the angle
ACE, the less to the greater,
which is impossible. Where-
fore F is not the centre of the

D

c
circle ABC. In the same man-
ner it
may

be shown that no other point which is not in CA is the centre; that is, the centre is in CA. Therefore, if a straight line, &c. Q. E. D.

PROP. XX. THEOR.

THE angle at the centre of a circle is double of the angle at the circumference, upon the same base, that is, upon the same part of the circumference.

Let ABC be a circle, and BDC an angle at the centre, and
BAC an angle at the circumference, which have the same
circumference BC for their base; the angle BDC is double
of the angle BAC.
First, let D, the centre of the

A
circle, be within the angle BAC,
and join AD, and produce it to
E. Because DA is equal to DB,
the angle DAB is equala to the

D
angle DBA; therefore the angles
DAB, DBA are double of the
angle DAB. But the angle BDE

BS
is equal to theangles DAB, DBA;
therefore the angle BDE is double

TE

a 5. 1.

b 32. 1.

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