straight line from D to the circumference, and DC is greater than DB, and DB greater than DA; but they are equal, by hypothesis; which is impossible. Therefore E is not the centre of ABC. In the same manner it may be shewn that any other point than D is not the centre; therefore D is the centre of the circle ABC.

X.-One circumference of a circle cannot cut another at more than two points.

If possible, let the circumference ABC cut the circumference DEF at more than two points, namely at B, G, F.

Take K, the centre, join KB, KG, KF. Because K is the centre, KB, KG, KF are all equal. And because in DEF, K is taken, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF, therefore K is

B

A

H

C

the centre of the circle DEF. But K is also the centre of the circle ABC. Therefore the same point is the centre of two circles which cut one another, which is impossible.

XI.-If two circles touch one another internally, the straight line which joins their centres, being produced, shall pass through the point of contact.

Let ABC, ADE touch one another internally at A; and let F be the centre of ABC, and G of ADE; the straight line which joins F, G, being produced, passes through A.

If not, let it pass otherwise, if possible, as FGDH, and join AF, AG. H/ Because AG, GF are greater than AF, and AF is equal to HF, AG, GF are greater than HF. the common part GF; therefore the remainder AG is greater than HG.

G

Take away

But AG is equal to DG. Therefore DG is greater than HG, the less than the greater; which is impossible. Therefore the straight line which joins F, G, being produced, cannot pass otherwise than through A, that is, it must pass through A.

XII.-If two circles touch one another externally, the straight line which joins their centres shall pass through the point of contact.

B

E

Let ABC, ADE touch one another externally at ▲, and let F be the centre of ABC, and G of ADE; the straight line which joins F, G, shall pass through A. If not, let it pass otherwise, if possible, as FCDG, and join FA, AG. Because F is the centre of ABC, FA is equal to FC; because G is the centre of ADE, GA is equal to GD; there

A

D

Therefore the

But FG is also
Therefore

fore FA, AG are equal to FC, DG. whole FG is greater than FA, AG. less than FA, AG; which is impossible. the straight line which joins F, G, cannot pass otherwise than through A, that is, it must pass through A. XIII.-One circle cannot touch another at more points

than one, whether it touches it on the inside or outside. For, if it be possible, let the circle EBF touch ABC at more points than one; and first on the inside, at B, Join BD, and draw GH bisecting BD at right angles. Because B, D, are in the circumference of each of the circles, BD falls within each;

D.

and therefore the centre of each circle is in GH, which bisects BD at right angles; therefore GH passes through the point of contact. But GH does

not pass through the point of contact, because B, D are out of GH; which is absurd.

A

K

Therefore one circle cannot touch another on the inside at more points than one. Nor can one circle touch another on the outside at more points than one. For, if possible, let ACK touch ABC at A, C. Join AC. Because A, C are in the circumference of ACK, AC, which joins them, falls in ACK; but ACK is without ABC, therefore AC is without ABC. But because A, C are in the circumference of ABC, AC falls within the B circle ABC; which is absurd. There

fore one circle cannot touch another on the outside at more points than one.

XIV.-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 AB, CD in ABDC, be equal to one another; they are equally distant from the centre.

Take E, the centre of ABDC; and from E draw EF, EG perpendiculars to AB, CD; and join EA, EC. Because EF, passing through the centre, cuts AB, which does not pass through the centre, at right angles, it also bisects it; therefore AF is equal to

B

FB, and AB is double of AF. For the like reason CD is double of CG. But AB is equal to CD; therefore AF is equal to CG. And because AE is equal to CE, the square on AE is equal to the square on CE. But the squares on AF, FÉ are equal to the square on AE, because AFE is a right angle; and the squares on CG, GE are equal to the square on CE; therefore the squares on AF, FE are equal to the squares on CG,

GE. But the square on AF is equal to the square on CG, because AF is equal to CG; therefore the remaining square on FE is equal to the square on GE; and EF to EG. But straight lines in a circle are equally distant from the centre, when the perpendiculars drawn to them from the centre are equal; therefore AB, CD are equally distant from the centre.

Next, let AB, CD be equally distant from the centre, AB is equal to CD. For, it may be shewn, as before, that AB is double of AF, and CD of CG, and the squares on EF, FA are equal to those on EG, GC; but the square on EF is equal to that on EG, because EF is equal to EG; therefore the remaining square on FA is equal to that on GC, and AF to CG. But AB is double of AF, and CD double of CG. Therefore

AB is equal to CD. XV.-The diameter is the greatest straight line in a circle; 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, AD the diameter, and E the centre; and let BC be nearer to the centre than FG; AD is greater than BC, which is not a diameter, and BC than FG. From F E draw EH, EK perpendiculars to BC, FG, and join EB, EC, EF. Because AE is equal to BE, and ED to EC, therefore AD is equal to BE, EC; but BE, EC are greater than BC; c therefore also AD is greater than BC.

F

A B

D

A B

H

And because BC is nearer to the centre than FG, EH is less than EK. Now BC is double of BH, and FG double of FK, and the squares on EH, HB are equal to the squares on EK, KF. But the square on EH is less than the square on EK, because EH is less than EK; therefore the square on HB is greater than on KF; and BH is greater than FK; and BC than FG. Next, let BC be greater

D

H

than FG; BC is nearer the centre than FG, or EH is less than EK. Because BC is greater than FG, BH is greater than FK. But the squares on BH, HE are equal to those on FK, KE; and the square on BH is greater than that on FK, because BH is greater than FK; therefore the square on HE is less than the square on KE; and therefore EH is less than the straight line EK.

XVI.-The straight line drawn at right angles to the diameter 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. Let ABC be a circle, D the centre, and AB a diameter: the straight line drawn at right angles to AB from A, falls without the circle.

If not, let it fall, if possible within the circle, as AC, and draw DC to C, where it meets the circumference. B Because DA is equal to DC, the angle DAC is equal to DCA. But DČA is a right angle; therefore DCA is

A

a right angle; and therefore DAC, DCA are equal to two right angles; which is impossible. Therefore tho straight line drawn from A at right angles to AB does not fall within the circle. In the same manner it may be shewn that it does not fall on the circumference. Therefore it must fall without the circle, as AE. Also between AE and the circumference, no straight line can be drawn from A, which does not eut the circle.

If possible, let AF be between them; and from D draw DG perpendicular to AF; let DG meet the cir- B cumference at H. Then, because DGA is a right angle, DAG is less than a right angle; therefore DA is

H

FE

greater than DĞ. But DA is equal to DH; there fore DH is greater than DG, the less than the greater;