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is equal to ED; therefore GF, FE are greater than ED: take away the common part FE, and the remainder GF is greater than the remainder +5 Ax. FD. Therefore, FĂ is the greatest, and FD the least of all the straight lines from F to the circumference; and BF is greater than CF, and CF than GF.

Also, there can be drawn only two equal straight lines from the point Fto the circumference, one upon each side of the shortest line FD. At the point E, in the straight line EF, make * * 23. 1. the angle FEH equal to the angle FEG, and join FH : then, because GE is equalt to EH, †15 Def. and EF common to the two triangles GEF, HEF; the two sides GE, EF are equal to the two HE, EF, each to each; and the angle GEF is equal † to the angle HEF; therefore,

the base f Const. FG is equal* to the base FH: but, besides FH, * *. I. no other straight line can be drawn from F to the circumference equal to FG: for, if there can, let it be FK : and, because FK is equal to FG, and FG to FH, FK is equal to FH; that +1 Ax. is, a line nearer to that which passes through the centre, is equal to one which is more remote; which has been proved to be impossible. Therefore, if any point be taken, &c. Q. E. D.

PROP. VIII. THEOR.
If any point be taken without a circle, and straight

lines be drawn from it to the circumference,
whereof one passes through the centre ; of those
which fall upon the concave circumference, the
greatest is that which passes through the centre,
and of the rest, that which is neurer to the one
passing through the centre is always greater
than one more remote : but of those which fall
upon the convex circumference, the least is that
between the point without the circle and the
diameter; and of the rest, that which is nearer
to the least is always less than one more remote :
and only two equal straight lines can be drawn

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*1. 3.

from the same point to the circumference, one upon each side of the least line.

Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC be drawn to the circumference, whereof DA passes through the centre. Of those which fall upon the concave part of the circumference AEFC, the greatest shall be DA, which passes through the centre; and the nearer to it shall be greater than the more remote, viz. DE greater than DF, and DF greater than DC: but of those which fall upon the convex circumference HLKG, the least shall be DG between the point D and the diameter AG; and the nearer to it shall be less than the more remote, viz. DK less than DL, and DL less than DH.

Take* M the centre of the circle ABC, and join ME, MF, MC, MK, ML, MH. And be

cause AM is equal to ME, add MD to each, + 2 Ax. therefore AD is equal + to EM, MD: but EM, *20. 1. MD are greater* than ED; therefore also AD

is greater than ED. Again, because ME is
equal to MF, and M) common to the triangles
EMD, FMD; EM, MD,
equal to FM, MD, each to

D
each : but the angle EMD is
19 Ax. greater | than the angle FMD;

therefore the base ED is * 24. 1. greater* than the base FD). AROBN

H
În like manner it may be
shown that FD is greater than
CD. Therefore, DA is the

M
greatest; and DE greater than

F
DF, and DF greater than DC.
And, because MK, KD are

E *0.1. greater * than MD, and MK is +15 Def. equalt to MG, the remainder KD is greater*

than the remainder GD, that is, GD is less than KD: and because MLD is a triangle, and from the points M,D, the extremities of its side MD,

are

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the straight lines MK, DK are drawn to the point K within the triangle, therefore MK, KD are less * than ML, LD: but MK is equal † to *21. 1.

therefore, the remainder DK is less than †15 Del the remainder DL. In like manner it may be !

75 Ax, shown, that DL is less than DH. Therefore, DG is the least, and DK less than DL, and DL less than DH. Also, there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least line. At the point M, in the straight line MD, maket the angle DMB equal to the angle DMK, † 23. 1. and join DB: and because MK is equal to MB, and MD common to the triangles KMD, BMD, the two sides KM, MD are equal to the two BM, MD, each to each; and the angle KMD is equal t to the angle BMD; therefore the base + Const. DK is equal* to the base DB : but, besides DB, * 4. 1. there can be no straight line drawn from D to the circumference equal to DK: for, if there can, let it be DN: and because DK is equal to DN, and also to DB, therefore DB is equal to DN; that is, a line nearer to the least is equal to one more remote, which has been proved to be impossible. If, therefore, any point, &c. Q.E.D.

PROP. IX. THEOR. If a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle.

Let the point D be taken within the circle ABC, from which to the circumference there fall more than two equal straight lines, viz. DA, DB, DC: the point D shall be the centre of the circle.

For, if not, let E he the centre : join DE, and produce it to the circumference in F, G; then FG is a diametert of the circle ABC: and be- †17 De£ cause in FG, the diameter of the circle ABC, there is taken the point D, which is not the

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*7.3.

centre, DG is the greatest line
from it to the circumference,
and DC is greater
* than DB, F

DE

G and DB greater than DA: but + Hyp. they are likewise t equal, which

is impossible : therefore E is
not the centre of the circle
ABC. In like manner it may be demonstrated,
that no other point but D is the centre; D
therefore is the centre. Wherefore, if a point
be taken, &c. Q.E. D.

PROP. X. THEOR.
One circumference of a circle cunnot cut another

in more than two points.
If it be possible, let the cir-
cumference FAB cut the cir-
cumference DEF in more than B D H

two points, viz. in B, G, F: E K *3.3.

take f the centre K of the circle
ABC, and join KB, KG, KF:
then because K is the centre of

the circle ABC, therefore KB, 115 Def. KG, KF are all equal to each other: and be

cause within the circle DEF there is taken the point K, from which to the circumference DEF fall more than two equal straight lines KB, KG,

KF, therefore the point K is * the centre of the f Const, circle DEF: but K is also the centre t of the

circle ABC; therefore the same point is the

centre of two circles that cut one another, which *5.3.

is impossible.* Therefore, one circumference of
a circle cannot cut another in more than two
points. Q. E. D.

PROP. XI. THEOR.
If two circles touch each other internally, the

straight line which joins their centres being
produced shall pass through the point of con-
taci.
Let the two circles ABC, ADE touch each

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other internally in the point A; and let F be
the centre of the circle ABC, and G the centre
of the circle ADE; the straight
line which joins the centres F,
G, being produced, shall pass HA
through the point A.

For, if not, let it fall otherwise, if possible, as FGDH, and E join AF, AG. Then, because B two sides of a triangle are together greatert than the third side, therefore FG, +20. 1. GA are greater than FA: but FA is equal+ to 15 Def FH; therefore FG, GA are greater than FH : 1. take away the common part FG; therefore the remainder AG is greater t than the remainder +5 Ax. GH; but AG is equal t to GD; therefore GD +15 Def. is greater than GH, the less than the greater, which is impossible. Therefore the straight line which joins the points F, G, being produced, cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, &c. Q E. D.

PROP. XII. THEOR.
If two circles touch each other externally, the

straight line which joins their centres shall
pass through the point of contact.

Let the two circles ABC, ADE, touch each other externally in the point A; and let F be the centre of the circle ABC, and G the centre of ADE: the straight line which joins the points F, G, shall pass through the point of contact A.

For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG. And because F is the centre of the circle ABC, FA is equal to

E

B
FC: also, because G is
the centre of the circle

Α.
ADE, GA is equal to
GD: therefore FA, AG
are equal t to FC, DG;

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