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tains the corresponding circular sector cno.

Let AD be the side of a similar

polygon inscribed in the circle ADB, and draw the radii AC, CD, ac, cd. The similar triangles ACD, acd, give AC: ac :: AD: ad, and :: perim. of polygon in ABD perim. of polygon in abd. But, by the preceding theorem, AC : ac :: circumf. ABD circumf. abd. The perimeters of the polygons are, therefore, as the circumferences of the circles. But this is impossible; because (hyp.) the perimeter of polygon in ABD is less than the circumference; while, on the contrary, the perimeter of polygon in adb is greater than the circumference ikk. Consequently, AC is not to ac, as circumference ADB, to a circumference less than adb. And by a similar process it may be shown, that ac is not to AC, as the circumference ahd, to a circumference less than ABD. Therefore AC: ac :: circumference ABD circumference abd.

Cor. If R, r be the radii, D, d the diameters, and C, c the circumferences, we have, by this theorem, C: c:: R : r ; or, if C = π R, c = πr; and, by the former, area (A) : area (a) :: †RC: rc: we have A: a :: R2 : 1⁄2πr2 :: R2 : r2 :: D2 : d2 :: C2 : c2.


Angles at the centre of a circle, angles at the circumference of a circle, and sectors of circles, have all the same ratio as the arcs by which they are subtended.

LET AB, BD be two arcs of a circle subtending the angles ACB, BCD at the centre, the angles AMB, BFD at the circumference, and the sectors: then,

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First. Take any number of arcs AL, LE, each equal to AB, and any number DG, GH, HK, each equal to BD, and draw CL, CE, CG, CH, CK.

Then since BA, AL, LE, are all equal, the angles ACB, LCA, ECL, are all equal (ax. 11), and hence whatever multiple the arc EB is of the arc AB, the same multiple is the angle ECB of the angle ACB. In like manner, whatever multiple the arc BK is of the arc BD, the same multiple is the angle KCB of the angle DCB.

Again, if the arc BE be greater than the arc BK, the angle ECB is greater than the angle BCK; if equal, equal; if less, less: and these are equimultiples of AB, and ACB the first and third, and of BD, BCD the second and fourth. Hence (th. 77) it follows that

angle ACB angle BCD :: arc AB : arc BD.

Secondly. The angles AMB, BFD, at the circumference being the halves of the angles ACB, BCD, respectively, at the centres, have the same ratios; that is,

angle AMB angle BFD :: angle ACB : angle BCD :: arc AB: arc BD. Thirdly. The sectors ECL, LCA, ACB, are equal, and also the sectors BCD, DCG, GCH, HCK, are equal. Conceive the sector ACB to be placed upon ECL, so that CB shall coincide with CL; then the angles ACB, ECL, being equal, as before, the side AC will coincide with CE. Then the arc AB will coincide with the arc LE. For if not, let it take some other position, as EPL, and draw CPQ, cutting the arcs in P and Q; then since PC and QC are radii of the same circle they are equal: whence it is impossible that the arcs AB, EL should not coincide. The sector ECL is therefore equal to the sector ACB. In

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like manner, the sector LCA is equal to ACB; and the sectors DCG, GCH, HCK, are each equal to the sector BCD. Hence it may be proved, as in the first case, that the sectors are to one another as the arcs on which they stand.


If the three sides of a triangle be cut by any straight line, any one side will be divided in a ratio compounded of the ratios of the segments of the other two.

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LET ABC be a triangle cut by any line, straight or transversal, in D, E, F: then


BF: FA :: CE. BD: AE. CD
AF. BD. CE = FB. DC. AE.

For, draw BH parallel to AC meeting EDF in H.

by sim. trian. HBF, AFE, AE: BH :: AF: FB, and

by sim. trian. HBD, EDC, BH: EC :: BD : DC;

hence, by composition, cancelling BH from the first and second terms,


The two next are obtained in a similar manner; and the last by th. 81, p. 230.


If three straight lines be drawn from the angles of a triangle through any point, to meet the opposite sides, the segments of any one side will be divided in a ratio compounded of the ratios of the segments of the other two.

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AE : EC :: AF . BD : FB . DC
CD: DB:: AF. EC : FB. AE
BF: FA:: CE. BD : AE. CD

LET P be any point, through which lines AP, BP, CP, from the angles of the triangle ABC are drawn to meet the sides in D, E, F, respectively: then | AF. BD. CE = FB. CD . EA. For, through B draw HG parallel to AC meeting AP, CP, in H and G. Then

by parallels AC, HG, cut by HP, BP, GP, AE : EC :: BH: BG, sim. trian. FAC, FBG, AC: BG:: FA: FB, and

sim. trian. DCA, DBH, HB: AC:: BD; DC;

hence, by composition of ratios, we get


Similarly the two next may be obtained; and the last as the corresponding one of the last theorem.


Though the sides of the triangles in these two propositions (th. 95, 96) have the same relation as to ratio amongst their segments, yet there is an essential difference as to the number of intersections made in the sides produced and unproduced. In the latter theorem, there are always two or none produced, but never one singly or all three; that is, an even number: in the former, always one or three, but never two or none; that is, an odd number. This distinction will appear to be of importance in the next theorem, which is the converse of the two last.


If the sides of a triangle be divided so that the segments of one side have the ratio compounded of the ratios of the segments of the other two sides, then :

1. If two or more of these sides be so divided in prolongation, lines drawn from the points of section to the opposite angles, will all pass through the same point.

2. If one or three of the sides be divided in prolongation, the three points of section will be in one straight line.

First. (figs. th. 96.) LET AE EC:: AF. BD: FB. DC, two sides or more being produced, then AD, BE, CF, pass through one point.

For if not, let CF, AD, intersect in P, and draw BP to meet AC in E'. Then (th. 96) AE' E'C :: AF. BD: FB. DC. Whence AE': E'C :: AE : EC, and AEEC : AE' :: AE ± EC: AE. But AE' EC = AE + EC; hence AE' = AE the less to the greater, which is impossible. The three lines AD, BE, CF, therefore pass through the same point P.

Second. (figs. th. 95.) LET AE : EC :: AF.BD : FB . DC, one side or three being produced; then D, E, F, are in one straight line. For if not, let FD meet AC in E'. AF. BD FB. DC, and hence AE' AE AE EC: AE; but AE' the less to the greater, which is impossible. Whence D, E, F, are in one line.

Then, (th. 95,) we have AE EC:: E'C:: AE: EC, and AE' E'C: 'EC = AE = EC, and hence AE' = AE,


This proposition furnishes a ready method of proving a great number of elegant theorems. For instance the following:

(1). The three perpendiculars from the angles of a triangle to the opposite sides, meet in one point.

(2). The lines which bisect the angles of a triangle, either all internally, or one internally and the other two externally, meet in one point.

(3). The lines which bisect the sides of a triangle meet in one point.

(4). Lines drawn from the angles of a triangle to the points of contact of its inscribed circle, meet in one point.

(5). Lines drawn from the angles of a triangle to the points of contact of circles each touching one side exteriorly and the other two produced, meet in one point.

(6). If the exterior angles of a triangle be bisected by lines which are produced to cut the opposite sides, the bisecting lines intersect those sides in three points which lie in one straight line.


If through any point lines be drawn from the angles of a triangle to cut the opposite sides, and the points of section be joined two and two by lines which are produced to cut the remaining sides of the triangle; then the sides of the triangle will all be divided in harmonical proportion, each in the two points in which it is cut by the lines described, and the last-mentioned three points of section will be in one straight line.

LET ABC be a triangle, from the angles A, B, C, of which lines are drawn through any point P to meet the opposite sides in D, E, F, respectively; and let EF be drawn to meet BC in D', FD to meet AC in E', and DE to meet AB in F': then First, BD DC :: BD': D'C, CE: EA: CE' : E'A,

and FA BF :: F'A : BF'.


For, (th. 96,) AE: EC:: AF. BD: FB. CD,
and, (th. 95,) AE': E'C:: AF. BD : FB. CD.

Whence AE EC:: AE': E'C; and similarly of the other two proportions. The lines are therefore divided harmonically (def. 84).

Second. The points, D', E', F', are in one straight line. For, compounding the three ratios, we have

BD.CE. FA; DC. EA . FB :: BD'. CE'. F'A: D'C. E'A. F'B; and since the first term is equal to the second, the third term is equal to the fourth, and CE : E'A :: F'B. CD': F'A . BD'. Hence (th. 97) the points D', E', F', are in one straight line.

Cor. 1. If from one angle, as B, of a triangle ABC, a line BE be drawn to cut the opposite side in E, and from the other angles any number of pairs of lines be drawn to meet in BE and cut the opposite sides, as AD, CF, meeting AE in P, AH, CG, meeting it in Q, and so on ; then DF, HG, and so on, will all meet AC in the same point E'. For they all divide, with E, the side AC harmonically in their points of intersection. These points, must, therefore, coincide.

Cor. 2. If from any point E' in one side of a triangle lines be drawn to cut the other two, as E'GH, E'FD, and so on; and lines be drawn from these points to the angles opposite, they will two and two cut each other in points, all of which lie in one straight line.

For (th. 97) BQ, BP, and so on, all cut the side AC in points such that with E' divide it harmonically. Hence these points must coincide, and therefore BQ, BP, .... must also coincide, or the points P, Q, .... all lie in one straight line*.

* Several simple and yet very important properties of harmonical lines are thrown together in this note, chiefly to avoid the necessity of formal enunciations, which, if given in words, would occupy considerable space.

1. Let AC CB:: AD: DB be the general division of the line. Bisect AB in M and CD in M'. Then

2. CA: AD:: CB: BD, or the line CD is harmonically divided in A and B; and the same points of division result from supposing either AB or CD to be the given line.


3. From


If a straight line be divided harmonically, and from the four points of section, straight lines be drawn through any point in the same plane: then

1. Any straight line drawn parallel to one of those four lines will be bisected by the

other three.

2. Any line oblique to them all will be harmonically divided by them at the points of intersection.

First. LET AC: CB :: AD: DB, and the lines be drawn from A, B, C, D, to meet at F: then, if a line MKL be drawn parallel to AF, one of the extreme lines of the fasceau, it will be bisected in K.

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Through B draw HBE parallel to ML or FA. Then, by parallels AD: DB:: AF: BH, and AC: CB:: AF : BE; but by hypothesis, AD: DB:: AC: CB, and therefore AF: BE:: AF: BH, or BE = BH. Again, by parallels, ML, HE, HB: BE:: MK: KL, or MK = KL.

Next, let the line L'K'M' be parallel to one of the intermediate lines FC, it will be bisected in M'.

Through B draw BI'H' parallel to L'K' or FC, cutting the extreme lines of the fasceau in I', H'; and draw HI through B parallel to AF.

Then, because BH' is parallel to FH, and BH to FH',







we have FH' = BH: but by the previous case, BH = BI; and hence FH' is equal and parallel to BI, and the diagonal H'B is bisected in I' by the diagonal FI, or extreme sector FD. Also, by parallels, H'I' : I'′B :: L'M' : M'K', and M'K'.

HI' = I'B; hence L'M'

Cor. Conversely, if from any point two lines be drawn to the extremities of a given line, a third to bisect that line, and a fourth parallel to it, they will form an harmonical fasceau.

Second. If any line cut an harmonical fasceau F{ABCD} it will be divided harmonically in the points A', B', C', D', of intersection.

For, through B' draw KI parallel to AF: then by the former part of the proposition, KI is bisected in B'; and by parallels

A'F: B'I :: A'D': B'D', and

A'F. B'K: A'C': C'B': whence



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A'D': D'B' :: A'C' : C'B', and A'B' is harmonically divided in C', D'.

3. From (1) we have ACCB; AC+ CB:: AD-DB: AD + DB; that is, 2MC: 2MB 2MB 2MD, or MC: MB:: MB: MD; or MB is a mean proportional between MC and MD.

4. In like manner, M'B: M'C:: M'C: M'A, or M'C is a mean proportional between M'A and M'B.

5. From (3) we have MD MD + MD:: MB: MB MC, or which is the same thing, MD: DB:: MB: BC.

6. By th. 33, AD. DB = MD2- MB2, and AC. CB = MB2

ence gives us at once

MC2; hence the differ

AD. DB-AC. CB MD2 2MB2 + MC2, or by (3)
= MD22MC. MD + MC2, or th. 32,
= (MD- MC)2 = CD2.

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