Sidebilder
PDF
ePub

Let the circle ABC be touched internally by the circle DEC in the point C: they have not the same centre.

F

For, if they can, let it be F; join FC, and draw any straight line FEB meeting the circumferences in E and B. Then (I. def. 30) because F is the centre of the circle ABC, FC is equal to FB; and, because F is the centre of the circle CDE, FC is equal to FE. But FC has been shown to be equal to FB: therefore FE is equal to FB, which (I. ax. 9) is impossible; wherefore F is not the centre of the circles ABC, CDE; and in the same manner it can be shown, that no other point can be the centre of both the circles. Therefore, if one circle, &c.

D

E B

PROP. VII. THEOR.-If from any point within a circle, which is not the centre, straight lines be drawn to the circumference: (1) the greatest is that which passes through the centre, and (2) the continuation of that line to the circumference, in the opposite direction, is the least: (3) of others, one nearer to the line passing through the centre is greater than one more remote: and (4) from the same point there can be drawn only two equal straight lines, one upon each side of either the longest or shortest line, and making equal angles with that line.

Let ABCD be a circle. E its centre, and AD a diameter, in which let any point F be taken, which is not the centre: of all the straight lines FA, FB, FC, &c., that can be drawn from F to the circumference, FA is the greatest, and FD the least: and of the others, FB is greater than FC.

1. Join BE, CE. Then (I. 20) BE, EF are greater than BF; but AE is equal to EB; therefore AF, that is, AE, EF, is greater than BF. 2. Because CF, FE are greater (I. 20) than EC, and EC is equal to ED; CF, FE are greater than ED. Take away the com. mon part FE, and (I. ax. 5) the remainder CF is greater than the remainder FD.

3. Again, because BE is equal to CE, and FE common to the triangles BEF, CEF; but the angle BEF is greater than CEF: therefore (I. 24) the base BF is greater than the base CF.

B

བ།

D

H

K

4. Make (I. 23) the angle FEH equal to FEC, and join FH. Then, because CE is equal to HE, EF common to the two triangles CEF, HEF, and the angle CEF equal to the angle HEF;" therefore (I. 4) the base FC is equal to the base FH, and the angle EFC to the angle EFH. But, besides FH, no other straight line can be drawn from F to the circumference equal to FC. For, if there can, let it be FK; and because FK is equal to FC, and FC to FH, FK is equal (I. ax. 1) to FH; that is, a line nearer to

that which passes through the centre, is equal to one which is more remote; which is impossible. Therefore, if from any point, &c.

PROP. VIII. THEOR.-If from any point without a circle straight lines be drawn to the circumference; (1) of those which fall upon the concave part of the circumference, the greatest is that which passes through the centre; and (2) of the rest, one nearer to the greatest is greater than one more remote. (3) But of those which fall upon the convex part, the least is that which when produced, passes through the centre; and (4) of the rest, one nearer to the least is less than one more remote. And (5) only two equal straight lines can be drawn from the point to either part of the circumference, one upon each side of the line passing through the centre, and making equal angles with it.

Let ABF be a circle, M its centre, and D any point without it, from which let the straight lines DA, DE, DF be drawn to the circumference. Of those which fall upon the concave part of the circumference AEF, the greatest is DMA, which passes through the centre; and a line DE nearer to it is greater than DF, one more remote. But of those which fall upon the convex circumference LKG, the least is DG, the external part of DMA; and a line DK nearer to it is less than DL, one more remote.

1. Join ME, MF, ML, MK; and because MA is equal to ME, add MD to each; therefore AD is equal to EM, MD: but (I. 20) EM, MD are together greater than ED; therefore also AD is greater than ED.

2. Because ME is equal to MF, and MD common to the triangles EMD, FMD, but the angle EMD is greater than FMD; therefore (I. 24) the base ED is greater than the base FD.

3. Because (I. 20) MK, KD are greater than MD, and MK is equal to MG, the remainder KD is greater (I. ax. 5) than the remainder GD; that is, GD is less than KD.

4. Because MK is equal to ML, and MD common to the triangles KMD, LMD, but the angle DMK less than DML; therefore the base DK is less (I. 24) than the base DL.

F

D

A

M

5. Make (I. 23) the angle DMB equal to DMK, and join DB. Then, because MK is equal to MB, MD common to the triangles KMD, BMD, and the angle KMD equal to BMD; therefore (I. 4) the base DK is equal to the base DB, and the angle MDK to the angle MDB. 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 equal to DN, and also to DB;

therefore DB is equal to DN; that is, a line nearer to the least equal to one more remote, which is impossible. If from any point, therefore, &c.

PROP. IX. THEOR.-A point in a circle from which more than two equal straight lines fall to the circumference, is the centre.

For, from any point which is not the centre, only two equal straight lines (III. 7) can be drawn to the circumference, and therefore a point from which more than two equal straight lines can be drawn to the circumference, cannot be any other than the centre. Therefore a point, &c.*

PROP. X. THEOR.-One circle cannot cut another in more than two points.

A

B

D

E

If it be possible, let the circle FAB cut the circle DEF in more than two points, viz. in B, G, F: take (III. 1) the centre H, of the circle ABC, and join HB, HG, HF. Then, because within the circle DEF there is taken the point H, from which there fall to the circumference DEF more than two equal straight lines HB, HG, HF, H is (III. 9) the centre of the circle DEF; but His also the centre of the circle ABC: therefore the same point is the centre of two circles that cut one another, which (III. 5) is impossible. Therefore one circle, &c.

F

C

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

*The following is an outline of a direct demonstration of this proposition: Let M in the diagram for the last proposition be a point from which the equal straight lines, MA, ME, MF, are drawn to the circumference. Then, if a straight line be drawn from M to the point of bisection of the chord joining AE, it will be perpendicular (hyp. and I. 8) to the chord; and, therefore (III. 1, cor.) the centre will be in that perpendicular. In a similar manner, it would be shown, that the centre would be in the straight line drawn from M to the point of bisection of the chord joining EF. Hence, therefore, M, the point common to the two perpendiculars, must be the centre.

+ Dr. Simson's enunciation is as follows:-"If two circles touch each other internally, the straight line which joins their centres, being produced, shall pass through the point of contact." This is liable to two objections. The circles do not "touch each other internally," the interior one being touched externally. The expression, "the point of contact," seems to assume that there is but one point of contact, which is proved in a following proposition, the 13th. Mr. Walker proposes to say, a point of contact; but this would seem to imply that there may be more points of contact than one. In the enunciation here given, these objections are obviated. A similar change is made in one part of the enunciation of the 12th proposition, and also of the 6th.

With regard to this proposition, since, till the 13th proposition is proved, we are not to assume that the circles touch one another in only one point, we ought to consider the case, in which, if possible, the continuation of FG would pass through

Let the circle ADE touch the circle ABC internally in the point A, and let F be the centre of ABC, and G the centre of ADE: the straight line which joins F and G, being produced, passes through A.

H L

For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG. Then, because (I. 20) AG, GF are greater than FA, that is, than FH, for FA is equal to FH, each being a radius of ABC; take away the common part FG: therefore (I. ax. 5) the remainder AG is greater than the remainder GH; but AG is equal (I. def. 30) to GD: therefore GD is greater than GH, which (I. ax. 9) is impossible. The straight line, therefore, which joins the points F, G cannot fall otherwise than upon A, that is, it must pass through it. Therefore, if one circle, &c.

B

E

C

PROP. XII. THEOR.-If two circles touch each other externally in any point, the straight line which joins their centres passes through that point.

Let the two circles ABC, ADE touch each other externally in a point A; aud let F and G be their centres: the straight line which joins F and G passes through A.

E

For, if not, let it pass otherwise, if possible, as FCDG, and join FA, ÁG. Then (I. def. 30) because F and G are the centres of the circles, AF is equal to FC, and AG to GD: therefore FA, AG are equal to FC, DG; wherefore the whole FG is greater than FA, AG, which (I. 20) is impossible: therefore the straight line which joins

B

the points F, G cannot pass otherwise than through the point A: it therefore passes through that point: wherefore, if two circles, &c.

PROP. XIII. THEOR.-One circle can touch another in only one point, whether it touch it internally or externally.

For, by the two preceding propositions, if two circles touch one another in any point, the straight line joining their centres passes through that point. If therefore the circles could touch one another in more points than one, their centres would lie in the straight line joining two such points. Now, that straight line would be a chord of each of the circles; and, therefore (III. 1, cor.) the centres would also lie in a straight line bisecting it, and perpendicular to

another point of contact, so that D and H would coincide. In this case, we should still have, as in the first part of the demonstration, AG greater than GH or GD; while, from the coincidence of D and H, GA and GD or GH must be equal. Similar remarks are applicable with regard to the next proposition.

it. Hence, through the centres there would pass two straight lines not coinciding, which (I. def. 3) is impossible. One circle, therefore, &c.

PROP. XIV. THEOR.-Equal chords in a circle are equally distant from the centre; and (2) chords which are equally distant from the centre are equal to one another.

Let the chords AB, CD, in the circle ABDC, be equal to one another; they are equally distant from the centre.

Take (III. 1) E the centre of the circle, and draw (I. 12) EF, EG perpendiculars to AB, CD: join also EA, EC. Then, because the straight line EF, passing through the centre, cuts the chord AB, which does not pass through the centre, at right angles, it also (III. 3) bisects it: wherefore AF is equal to FB, and AB is double of AF. For the same reason, CD is double of CG; but AB is equal to CD; therefore AF is equal (I. ax. 7) to CG. Then, in the right-angled triangles EFÁ, EGC, the sides EA, AF are equal to the sides EC, CG, each to each; therefore (I. 47, cor. 5) the sides EF, EG are equal. But chords in a circle are said (III. def. 3) to be 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.

B

[ocr errors]

D

Next, if the chords AB, CD be equally distant from the centre, that is, if FE be equal to EG, AB is equal to CD. For, the same construction being made, it may, as before, be demonstrated, that AB is double of AF, and CD of CG; and because the right-angled triangles EFA, EGC have the sides AE, EF equal to CE, EG, each to each, the sides AF, CG are also (I. 47, cor. 5) equal to one another. But AB is double of AF, and CD of CG; wherefore AB is equal (I. ax. 6) to CD. Therefore equal chords, &c.

PROP. XV. THEOR.-The diameter of a circle is the greatest chord: (2) of others, one nearer to the centre is greater than one more remote; and (3) 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 BC be nearer to the centre than FG; AD 15 greater than BC, and BC than FG.

1. From the centre draw (I. 12) EH, EK perpendiculars to BC, FG, and join EB, EC, EF. Then, because AE is equal to EB, and ED to EC, AD is equal to EB, EC: but EB, EC are greater (1. 20) than BC; wherefore also AD is greater than BC.

2. Again, because BC is nearer to the centre than FG, EH (III. def. 4) is less than EK;

F

K

E

AB

and the hypotenuses EB, ÉF are equal. Therefore, (I. 47, cor. C)

« ForrigeFortsett »