DEMONSTRATION

Because two sides of a triangle are greater than the third,

therefore BE, EF, are greater than BF; (1. 20)

but AE is equal to EB; (1. def. 15)

therefore AE, EF, that is AF, is greater than BF.

Again, in the triangles BEF, CEF,

because BE is equal to CE, and EF common to both,

the two sides BE, EF, are equal to the two CE, EF, each to each; but the angle BEF is greater than the angle CEF; (ax. 9)

therefore the base BF is greater than the base CF; (1. 24)

for the same reason, CF is greater than GF.

Again, because GF, FE, are greater than EG, (I. 20) and EG 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 FD; (ax. 5)

therefore FA is the greatest of all the straight lines from F to the circumference,

and FD is the least;

and BF is greater than CF, and CF than GF.

Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD.

CONSTRUCTION

At the point E, in the straight line EF, make the angle FEH equal to the angle GEF, (1. 23) and join FH.

DEMONSTRATION

Then in the triangles GEF, HEF,

because GE is equal to EH, (1. def. 15) and EF common to both, 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; (constr.)

therefore the base FG is equal to the base FH; (1. 4)

but besides FH, no 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 assumed to be equal to FG, and FG is equal to FH, therefore FK must be equal to FH; (ax. 1)

that 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.

THEOREM.

If any point be taken without a circle, and straight lines be drawn from it to the circumference, whereof one passes through the centre;

then of those which fall upon the concave circumference, the greatest is that which passes through the centre; and of the rest, that which is nearer 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 from the same point to the circumference, one upon each side of the line which passes through the

centre.

(References-Prop. I. 4, 20, 21, 23, 24; III. 1.)

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.

Then, of those which fall upon the concave part of the circumference AEFC, AD which passes through the centre shall be the greatest; and the nearer to it is always greater than the more remote, viz., DE than DF, and DF 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 is always less than the more remote, viz., DK than DL, and DL than DH.

D

GEN

M

CONSTRUCTION

Take M the centre of the circle ABC, (ш. 1) and join ME, MF, MC, MK, ML, MH.

DEMONSTRATION

And because AM is equal to EM,

add MD to each,

therefore AD is equal to EM, MD; (ax. 2)

but EM, MD, are greater than ED; (1. 20) therefore also AD is greater than ED.

Again, in the triangles EMD, FMD,

because ME is equal to MF, and MD common to both, EM, MD, are equal to FM, MD, each to each,

but the angle EMD is greater than the angle FMD; (ax. 9) therefore the base ED is greater than the base FD. (1. 24.) In like manner it may be shown that FD is greater than CD; therefore DA is the greatest line,

and DE is greater than DF, and DF than DC

And because MK, KD, are greater than MD, (1. 20)

and MK is equal to MG, (1. def. 15)

the remainder KD is greater than the remainder GD; (ax. 5)

that is, GD is less than KD ;

and because MK, DK, are drawn to the point K within the triangle MLD, from M, D, the extremities of its side MD,

therefore MK, KD are less than ML, LD; (1. 21)

but MK is equal to ML; (1. def. 15)

therefore the remainder DK is less than the remainder DL. (ax. 5.)

In like manner it may be shown that DL is less than DH;

therefore DG is the least, and DK less than DL, and DL 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 line through the

centre.

CONSTRUCTION

At the point M, in the straight line MD, make the angle DMB equal to the angle DMK, (1. 23) and join DB.

DEMONSTRATION

Then, in the triangles KMD, BMD,

because MK is equal to MB, and MD common to both,

the two sides KM, MD, are equal to the two BM, MD, each to each; and the angle KMD is equal to the angle BMD; (constr.)

therefore the base DK is equal to the base DB; (I. 4)

but besides DB 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 assumed to be equal to DN, and DK has been shown to be equal to DB,

therefore DB must be 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.-THEOREM.

If a point be taken within a circle, from which there can be drawn more than two equal straight lines to the circumference; then that point is the centre of the circle.

(References-III. 7.)

Let the point D be taken within the circle ABC, from which to the circumference there can be drawn more than two equal straight lines, viz. DA, DB, DC.

Then the point D shall be the centre of the circle.

For, if not, let E be the centre; join DE, and produce it to the circumference in FG.

DEMONSTRATION

Then FG is a diameter of the circle ABC; (1. def. 17)

and because in FG, the diameter of the circle ABC, there is taken the point D, not the centre,

therefore DG must be the greatest line from the point D to the circumference, and DC greater than DB, and DB than DA; (III. 7) ·

but these lines are likewise equal, (hyp.) 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;

therefore D is the centre of the circle ABC.

Wherefore, if a point be taken, &c.

Q. E. D.

PROP. X.- THEOREM.

One circumference of a circle cannot cut another in more than two points.

(References-III. 3, 5, 9.)

Let the two circles ABC, DEF, intersect one another.

Then their circumferences cannot cut each other in more than two points.

If it be possible, let the circumference FAB cut the circumference DEF in more than two points, viz. in B, G,

F;

take the centre K of the circle ABC, (III. 3) and join KB, KG, KF.

DEMONSTRATION

And because 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,

the point K must be the centre of the circle DEF, (III. 9)

but K is also the centre of the circle ABC;

therefore the point K must be the centre of two circles that cut one another,

which is impossible. (III. 5.)

Therefore one circumference of a circle cannot cut an

other in more than two points.

Q. E. D.