from F to the circumference, FA, that in which the centre is, shall be the greatest;

and FD, the other part of the diameter AD, shall be the least;

and of the rest, FB, the nearer to FA, shall be greater than FC, the more remote, and FC greater than FG.

Construction. Join BE, CE, GE.

Demonstration. Because two sides of a triangle are greater than the third side (I. 20); therefore

1. BE, EF are greater than BF;

but AE is equal to BE (I. Def. 15); therefore

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

Again, because BE is equal to CE, and FE common to the triangles BEF, CEF, 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

3. The base BF is greater than the base CF (I. 24).

For the same reason,

4. CF is greater than GF.

Again, because GF, FE are greater than EG (I. 20), and EG is equal to ED; therefore

5. GF, FE are greater than ED;

take away the common part FE, and

6. The remainder GF is greater than the remainder FD (Ax. 5).

Therefore, FA 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 F to the circumference, one upon each side of the diameter.

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

Demonstration. Then, because GE is equal to EH (I. Def. 15), 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 (constr.); therefore

but

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

2. Besides FH, no other straight line can be drawn from F to the circumference equal to FG;

for, if possible, let it be FK; and because FK is equal to FG, and FG to FH, therefore

3. FK is 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.

PROPOSITION 8.-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; of those which fall upon the concave part of the 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 part of the 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.

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, the greatest shall be DA, which passes through the centre; and any line nearer to it shall be greater than one more remote, viz., DE shall be greater than DF, and DF greater than DC; but of those which fall upon the convex part of the circumference HLKG, the least shall be

DG between the point D and the diameter AG; and any line nearer to it shall be less than one more remote, viz., DK less than DL, and DL less than DH.

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

Demonstration. And because AM is equal to ME, add MD to ach of these equals; therefore

1. AD is equal to EM, MD ( Ax. 2) ;

but EM, MD are greater than ED (I. 20); therefore also

2. AD is greater than ED.

Again, because ME is equal to MF, and MD common to the triangles EMD, FMD; EM, MD, are equal to FM, MD, each to each; but the angle EMD is greater than the angle FMD (Ax. 9); therefore

3. The base ED is greater than the base FD (I. 24).

In like manner it may be shown, that

4. FD is greater than CD.

Therefore DA is the greatest; and DE greater than DF, and DF greater than DC.

And, because MK, KD are greater than MD (I. 20), and MK is equal to MG (I. Def. 15), the remainder KD is greater than the remainder GD (Ax. 5); that is,

1. GD is less than KD;

and because MLD is a triangle, and from the points M, D, the extremities of its side MD, the straight lines MK, DK are drawn to the point K within the triangle; therefore

2. MK, KD are less than ML, LD (I. 21);

but MK is equal to ML (I. Def. 15); therefore

3. The remainder DK is less than the remainder DL (Ax. 5).

In like manner it may be shown, that

4. 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 line which passes through the centre.

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

Demonstration. And because MK is equal to MB, and MD common to the triangles KMD, BMD; the two sides KM, MD are equal

G

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

but

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

2. Besides DB, no straight line equal to DK can be drawn from D to the circumference;

for, if possible, let it be DN; and because DK is equal to DN, and also to DB, therefore

3. 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.

PROPOSITION 9.-Theorem.

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.

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

Construction. For, if not, let E, if possible, be the centre; join DE, and produce it to meet the circumference in F, G.

Demonstration. Then FG is a diameter of the circle ABC (I. Def. 17); and because in FG, the diameter of the circle ABC, there is taken the point D, which is not the centre, therefore

D E

1. DG is the greatest line drawn from it to the circumference, and DC is greater than DB, and DB greater than DA

(III. 7) ;

but these lines are likewise equal (hyp.), which is impossible; therefore

2. 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

3. D is the centre of the circle ABC.

Wherefore, if a point be taken, &c. Q.E.D.

PROPOSITION 10.-Theorem.

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

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

Then their circumferences cannot cut each other in more than two

points.