is greater than ED. Again, because ME is equal to MF, and MD common to the two triangles EMD, FMD; the two sides EM and MD are equal to the two sides FM and M D, each to each. But the angle EMD is greater than (I. Ax. 9) the angle FMD. Therefore the base ED is greater than (I. 24) the base FD. In like manner, it may be shown that FD is greater than CD. Therefore, DA is the greatest; DE is greater than DF, and DF greater than DC.

Because MK and KD are greater than MD (I. 20) and MK is equal (I. Def. 15) to MG. Therefore the remainder KD is greater than (1. Ax. 5) the remainder GD; that is, GD is less than KD. Again, because MI is equal to M K, and MD common to the two triangles MID, MKD, the two sides IM and MD are equal to the two sides KM and MD, each to each. But the angle IMD is greater than the angle KMD. Therefore, the base ID is greater than (I. 24) the base KĎ; that is, KD is less than ID. In like manner it may be shown, that DI, is less than DH. Therefore, DG is the least; DK is less than DI, and DI less than DH.

Lastly, at the point M, in the straight line MD, make the angle DMB equal to the angle D MK (I. 23), and join DB.

Because M K is equal to MB, and MD common to the triangles KMD, BMD, the two sides K M and MD are equal to the two sides BM and MD, each to each. But the angle K MD is equal (Const.) to the angle BMD. Therefore the base DK is equal (I. 4) to the base DB. And no straight line drawn from D to the circumference, but DB, can be equal to DK: for, if possible, let DN be equal to DK. Because DN is equal to DK, and it has been shown that DB is equal to DK. Therefore DB is equal (I. Ax. 1) 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.

Corollary 1.-If two secants of a circle, drawn from the same point without it make equal angles with a secant passing through that point and the centre, they are equal.

Corollary 2.-If two secants of a circle, drawn from the same point without it, make unequal angles with a secant passing through that point and the centre; that which makes the least angle with it, is the greatest.

PROP. IX. THEOREM.

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

Let D be a point within the circle ABC, from which the equal straight lines DA, DB, and DC are drawn to the circumference. The point D is the centre of the circle. For if the point D be not the centre, let the point E be the centre. Join D E, and produce it to meet the circumference in the points F and G (I. Def. 17).

Because FG is a diameter of the circle ABC, and from the point D, which is not the centre, straight lines DG, DC, DB, and DA are drawn to the circumference. Therefore D G is the greatest, 1) C is

DE

greater than DB, and DB greater than DA (III. 7). But they arc

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

The preceding demonstration depends on the supposition that the three straight lines DA, DB, and DC are all on one side of the diameter FG. But the point E might be so chosen, that DC and DB should be on opposite sides of the diameter, and then DC might be equal to DB instead of being greater than it. There is consequently a defect in this demonstration. The following demonstration, which is free from this defect, is generally adopted instead of it: Because, from any point which is not the centre, only two equal straight lines (III. 7) can be drawn to the circumference. Therefore a point from which more than two equal straight lines can be drawn to the circumference, is the centre. Wherefore the point D is the centre of the circle ABC.

Exercise. Give a direct demonstration of this proposition, founded on the corollary to Prop. III. of this Book.

PROP. X. THEOREM.

If two circles intersect one another, the circumference of the one cannot cut that of the other in more than two points.

Let ABC and DEF, be two circles which intersect one another. Their circumferences cannot cut each other in more than two points. If it be possible, let the circumference of the circle FAB cut the circumference of the circle DEF in more than two points,-viz., in B, G, and F. Take the centre K of the circle ABC (III. 3), and join KB, KG, and K F.

Because K is the centre of the circle ABC. The straight lines K B, K G and K F are all equal (I. Def. 15) to each other. Because from the point K, within

A

B

D

K

the circle DEF, more than two equal straight lines KB, KG, and KF are drawn to the circumference DEF. Therefore the point K is the centre (III. 9) of the circle DEF. But K is also (Const.) the centre of the circle ABC. Therefore the same point K is the centre of the two circles ABC and DEF that cut one another; which (III. 5) is impossible. Therefore, one circumference of a circle cannot cut another in more than two points. Q. E. D.

Exercise. If two circles intersect one another, the straight line which joins their centres, bisects their common chord at right angles.

PROP. XI. THEOREM.

If one circle touch another internally in any point, the straight line which joins their centres being produced passes through that point.

Let the circle A D E touch the circle A B C internally in the point A. The straight line which joins their centres being produced passes through the point A.

For if the straight line joining their centres do not pass through the point A, let it pass otherwise, if possible, as FGDH; let F be the centre of the circle ABC, and G the centre of the circle ADE. Join AF and AG.

A

Because two sides of a triangle AGF are together greater than (I. 20) the third side. Therefore FG and GA are greater than F A. But FA is equal (I. Def. 15) to FH. Therefore FG and GA are greater than FH. From these unequals, take away the common part FG. Therefore the remainder AG is greater than (I. Ax. 5) the remainder GH. But A G is equal (I. Def. 15) to GD. Therefore GD is greater than GH; that is, a part is greater than the whole; which is impossible. Therefore the straight line which joins the centres of the circles ADE and ABC, being produced, cannot pass otherwise than through the point A; that is, it must pass through the point A. Therefore, if one circle, &c. Q. E. D.

H

&

E

B

The demonstration of this proposition might be rendered clearer by changing the enunciation as follows:-If one circle touch another internally in any point, that point and the centres of the two circles are in the same straight line. This will form an exercise for the student.

PROP. XII. THEOREM.

If two circles touch each other externally in any point, the straight line which joins their centres shall pass through that point.

Let the two circles ABC, ADE touch each other externally in the point A. The straight line which joins their centres passes through the point of contact A.

For if the straight line joining their centres do not pass through the point A, let it pass otherwise, if possible, as FCDG; let F be the centre of the circle BAC, and G the centre of the circle EAD; and join FA and A G.

Because is the centre of the circle ABC (I. Def. 15), FA is equal to F C. Because G is

E

B

the centre of the circle ADE, GA is equal to GD. Therefore FA and AG are together equal (I. Ax. 2) to FC and D G ; and the whole F G is greater (I. Ax. 9) than FA and AG. But FAG is a triangle, and FG is also less than FA and AG (I. 20); which is impossible. Therefore the straight line which joins the centres of the circles BAC and EAD cannot pass otherwise than through the point A; that is, it must pass through the point A. Therefore, if two circles, &c. Q. E. D. The demonstration of this proposition might be improved like the preceding one, by making a similar change on its enunciation as follows:-If two circles touch each other externally in any point, that point and the centres of the two circles are in the same straight line. This will form another exercise for the student.

PROP. XIII. THEOREM.

One circle cannot touch another in more points than one, either internally or

externally.

Let the circle EBF touch the circle A B C internally in the point B. EBF cannot touch ABC in any other point.

G

B

H B

F

H
F

Join

For if it be possible, let E B F touch A B C in another point D. BD, and draw G H bisecting (I. 11) BD at right angles. Because the points B and D are in the circumference of both circles, the straight line BD lies (III. 2) within each of them. Therefore their centres are (III. Cor. 1) in the straight line G H which bisects BD at right angles. Because GH joins their centres it passes through (III. 11) the points of contact B and D. Therefore G H coincides with DB. But GH is also at right angles to BD (Const.); which is impossible. Therefore one circle EBF cannot touch another A B C, internally, in more points than one.

G

Again, let the circle A CK touch the circle ABC externally in the point A. A CK cannot touch A B C in any other point. For, if it be possible let ACK touch A B C in another point C. Join A C.

B

Because the two points A and C are in the circumference of the circle A CK, the straight line AC which joins them, lies within (III. 2) the circle ACK. But the circle ACK is without (Hyp.) the circle ABC. Therefore the straight line AC is without the circle ABC. Because the points A and C are in the circumference of the circle A B C, the straight line AC lies (III. 2) within the circle ABC. But it has been proved that AC also lies without the circle ABC; which is absurd. Therefore one circle ACK cannot touch another circle A B C, externally, in more points than one. Therefore, one circle, &c. Q. E. D.

The demonstration of this proposition might be abridged thus:-If one circie touches another in two points, the straight line which joins the points of contact is within both circles (III. 2). Therefore the centres of the circles are both in the straight line which bisects this common chord at right angles (III. 1 Cor.). But the straight line which joins their centres passes also through the points of contact (III. 11 and 12). Therefore the straight line, in which their centres are, bisects the common chord at right angles, and at the same time passes through its two extremities; which is impossible. Therefore, &c.

PROP. XIV. THEOREM.

Equal straight lines in a circle are equally distant from the centre; and conversely, straight lines equally distant from the centre, are equal to one another.

Let the straight lines A B and CD, in the circle BA CD, be equal to one another. They are equally distant from the centre.

Take E the centre of the circle ABDC (III. 1), and from E draw EF and EG perpendiculars to AB and CD (I. 12) respectively. Join E A and EC.

Because the straight line EF, passing through the centre, cuts the straight line AB, which does not pass through the centre, at right angles (III. 3), AB is bisected at F. Therefore AF is equal to FB, and AB

F

B

E

double of A F. For the same reason, CD is double of CG. But AB is equal (Hyp.) to CD. Therefore A F is equal (I. Ax. 7) to C G, and their squares are equal. Because AE is equal to EC (I. Def. 15), the square of A E is equal to the square of E C. But the squares of AF and FE are equal (I. 47) to the square of AE. Also, the squares of EG and GC are equal (I. 47) to the square of E C. Therefore the squares of AF and FE are equal (I. Ax. 1) to the squares of CG and GE. From these equals, take the squares of AF and CG, which were shown to be equal. Therefore the remaining square of EF is equal (I. Ax. 3) to the remaining square of EG, and the straight line EF to the straight line EG. But straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal (III. Def. 4). Therefore A B and CD are equally distant from

the centre.

Next, let the straight lines AB and CD be equally distant from the centre (III. Def. 4); that is, let FE be equal to EG. The straight lines A. B and CD are equal.

For, the same construction being made, it may, as before, be demonstrated, that AB is double of AF, and CD double of CG, that the squares of FE and EG are equal, and that the squares of EF and FA are equal to the squares of EG and GC. From these equals, take the squares of FE and EG, which are equal. Therefore the remaining square of AF is equal (I. Ax. 3) to the remaining square of CG; and the straight line AF to the straight line CG. But A B was shown to be double of A F, and CD double of CG. Therefore A B is equal (I. Ax. 6) to CD. Therefore equal straight lines, &c. Q. E.D.

Exercise. If any number of equal chords in a circle, be bisected, one circle passes through all the points of bisection and touches them at these points,

PROP. XV. THEOREM.

The diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is 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 B C be nearer to the centre than FG. The diameter AD is greater than any other straight line B C, and BC is greater than FG. From the centre E draw EH and EK perpendiculars to BC and FG (I. 12), respectively, and join EB, EC, and EF.

F

E

АБ

Because AE is equal (I. Def. 15) to EB, and ED to E C. Therefore AD is equal (I. Ax. 2) to EB and EC; but EB and EC are greater (I. 20) than B C. Therefore also AD is greater than BC. Because BC is nearer (Hyp.) to the centre than FG, EH is less (III. Def. 5) than EK. Therefore the square of EH is less than the square of E K. But, it may be shown as in the preceding proposition, that BC is double of BH, and FG cuble of FK, and that the squares of EH and HB are equal to the squares of EK and KF. But the square of EH is less than the square of EK. Therefore the square of B His greater than the square of FK, and the straight line BH greater than the straight line FK. Therefore also, B C is greater than F G.