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• 2. 3. + Hyp.

2. 3.

falls within * the circle ACK: but the circle ACK is with-
out t the circle ABC; 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
within the same circle ; which is absurd :
therefore one circle cannot touch another on

B
the outside 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.

PROPOSITION XIV.

с

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 shall be equally distant from the

centre. + 1.3. Taket E, the centre of the circle ABDC, and from it t, † 12. 1.

draw EF, EG, perpendiculars to AB, CD, and join EA, EC. Then, because the straight line EF, passing through the cen

tre, cuts the straight line AB, which does not pass through • 3. 3. the centre, at right angles, it also bisects * it; therefore AF

is equal to FB, and AB double of AF; for the same reason, + Hyp. CD is double of CG: but AB is equalt to CD; †7 Ax.

therefore AF is equal t to CG. And because + 15 Def. I. AE is equal † to EC, the square of AE is equal

to the square of EC: but the squares of AF, • 47. 1. FE are equal * to the square of AE, because

B the angle AFE is a right angle; and for the

like reason, the squares of EG, GC are equal to the square of + 1 Ax.

EC; 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 EF 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 * 4 Def. 3. centre are * equal; therefore AB, CD are equally distant from

the centre.

A

D

+ 3 Ax.

... Next, let the straight lines AB, CD be equally distant from the centre, that is † let FE be equal to EG: AB shall † 4 Def. 3, be 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: but the square of FE is equal to the square of EG, because FE is equal t to EG; + Hyp. therefore the remaining square of AF is equal † to the re- + 3 Ax. maining square of CG; and the straight line AF is therefore equal to CG: but AB was shewn to be double of AF, and CD double of CG; wherefore AB is equal t to CD. Therefore, + 6 Ax. equal straight lines, &c.

Q. E. D.

PROPOSITION XV.

Theor.The diameter is the greatest straight line in a circle; See N.

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 shall be greater than any straight line BC, which is not a diameter, and BC shall be greater

AB than FG.

F From the centre, draw + EH, EK perpen

H + 12. 1. diculars to BC, FG, and join EB, EC, EF : and because AE is equal † to EB, and ED to EC,

+ 15 Def. 1. therefore AD is equal t to EB, EC: but EB, EC

# 2 Ax. are greater * than BC; wherefore also, AD is greater than BC. * 20. 1.

And, because BC is nearer t to the centre than FG, EH is + Hyp. less * than EK: but, as was demonstrated in the preceding, * 5 Def. 3. BC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares EK, KF: 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 FK, and the straight line BH greater than FK, and therefore BC is greater than FG.

Next, let BC be greater than FG : BC shall be nearer to the centre than FG, that is t, the same construction being + 5 Def. 3. made, EH shall be less than EK. Because BC is greater than FG, BH likewise is greater than KF; and the squares of BH,

square of

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; and therefore BC is nearer + to the centre than FG. Wherefore, the dia meter, &c.

75 Def. 3.

Q. E. D.

PROPOSITION XVI.

See N.

THEOR.--The straight line drawn at right angles to the dia

meter 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 ; 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.

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. 5. I.

* 17. 1.

Let ABC be a circle, the centre of which is D, and the dia. meter 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 circumfer+ 15 Def. 1. ence. And because DA is equal t to DC, B В

А the angle DAC is equal * to the angle ACD: + Hyp.

but DAC is a right + angle; therefore ACD
is a right angle; and therefore the angles DAC, ACD are
equal to two right angles ; which is * impossible : therefore
the straight line drawn from A, at right angles to BA, does
not fall within the circle. In the same manner it may

demonstrated, that it does not fall upon the circumference ; * See fig. 2. 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: from the point D draw * DG perpendicular to AF, and let it

meet the circumference in H. And because AGD is a right • 17. 1.

angle, and DAG less * than a right angle, DA is greater * than

DG : but DA is equal t to DH ; therefore DH is greater than + 15 Def. 1.

DG, the less than the greater ; which is impossible : there

be

* 12. 1.

* 19. 1.

FE

fore no straight line can be drawn from the point A, between
AE and the circumference, which does not cut the circle ; or,
which amounts to the same thing, however
great an acute angle a straight line makes
with the diameter at the point A, or how-
ever small an angle it makes with AE, the BI

A
circumference must pass 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 angle.” · Q. E. D.

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 + 2 Def. 3. 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 2. 3. there can be but one straight line which touches the circle in the same point."

PROPOSITION 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.

Find * the centre E of the circle, and join AE; and from * 1. 3. the centre E, at the distance EA, describe the circle AFG; from the point D, draw DF at right angles to EA, and join * 11. 1. EBF, AB: AB shall touch 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 AE, + 15 Def. I. EB are equal to the two FE, ED, each to each ; and they contain the angle at E common to the two triangles AEB, FED;

H CE

* 4. 1. therefore the base DF is equal * to the base AB, and the triangle FED to the triangle AEB, and the other angles to the other angles ; therefore the angle EBA is equal to the angle EDF:

+ Constr. but EDF is a right † angle, wherefore EBA is a right t angle; + 1 Ax.

and EB is drawn from the centre: but a straight line drain

from the extremity of a diameter, at right angles to it, • Cor. 16.3. touches * the circle ; therefore, AB touches the circle ; 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 Cor. 16,3. right angles to DE: DF touches * the circle.

PROPOSITION XVIII.

THEOR.-If a straight line touches a circle, the straight line

drawn from the centre to the point of contact, shall be perpendicular to the line touching the circle.

an

Let the straight line DE touch the circle ABC in the point 1. 3.

C; take the centre F, and draw the straight line FC: FC shall be perpendicular to DE.

For if it be not, from the point F, draw FBG perpendi* 17. 1. cular to DE : and because FGC is a right angle, GCF is • 19. 1. acute angle ; and to the greater angle the greater * side is

opposite ; therefore FC is greater than FG : + 15 Def. 1. but FC is equal + to FB; therefore FB is

greater than FG, the less than the greater ; F
which is impossible; therefore FG is not
perpendicular to DE. In the same manner

B

D it may be shewn, that no other is perpendi

C GE cular to it besides FC; that is, FC is perpendicular to DE. Therefore, if a straight line, &c.

Q. E. D.

PROPOSITION 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 shall be in that line.

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 shall be in CA.

For, if not, let F be the centre, if possible,
and join CF. Because DE touches the circle BI
ABC, and FC is drawn from the centre to
the point of contact, FC is perpendicular

D

E to DE; therefore E is a right angle: but ACE is also a right t angle; therefore the angle FCE is

• 18. 3.

+ Hyp.

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