tion, then the three lines drawn from the opposite angles and forming these segments pass through the same point. That is, if AA' × BB' × CC' = AC' × CB' × BA', then the three lines AB', BC', and CA' pass through the same point.

T

Demonstration.-Let AB' and CA' be two of those lines intersect

CF=AF× CB' × BA', and hence
and produce it to cut AC in C';
× CC'
AC' × CB' × BA'.

AA' × BB'

Hence

Where

CB' × BA'

and we have AF: AC'::CF:CC'. But AF is

less than AC', being a part of it; hence CF must be less than its part CC', which is absurd. Whence the given relation can exist only when BF coincides with BEC', and the three lines pass through the same point. Q. E. D.

Scholium 1.-When the three lines AB', BC', and CA' bisect the three angles of a triangle respectively, they pass through the same point.

Demonstration.—Let the three lines AB', BC′, and CA' bisect the angles A, B, and C respectively; then we have (IV. 17*)

AC' : CC'::AB:BC.

CB': BB'::AC:AB.

BA': AA'::BC: AC.

Whence, by multiplying the corresponding terms of these proportions, we have AC' × CB' × BA' : CC' × BB' × AA' :: AB × AC × BC: AB × AC × BC::1:1. Hence AC' × CB' × BA' CC' × BB' x AA', and, by the theorem, the three lines pass through the same point.

Scholium 2.-When the three lines AB', BC', and CA' are perpendicular to the opposite sides respectively, they pass through the same point.

* VI. 3.

Demonstration.-In similar right-angled triangles ACB and AA'C, we have AC': AA'::BC' : CA'.

In similar right-angled triangles BA'C and BB'A, we have BA': BB':: CA' : AB'.

In similar right-angled triangles CB'A and CC'B, we have CB': CC:: AB': BC'.

Multiplying these proportions, we have AC' × CB' × BA': AA' × BB' x CC':: AB′ × CA' × BC' : AB' × CA' × BC'::1:1. Hence AA' × BB' × CC' AC' × CB' x BA', and the three lines pass

through the same point. (See Prob. XXVI., "Triangles," etc.)

Scholium 3.-When the three lines AB', BC', and CA' bisect the opposite sides respectively, they pass through the same point. (See Prob. XI., "Triangles," etc.)

Demonstration.-We here have AA' — BA', BB' BA', BB' = C'B', and CC' AC'; hence, by multiplying these three equations together, we have AA' × BB' × CC' — AC' × CB' × BA'; wherefore, by the theorem, the three lines pass through the same point. Theorem I.)

(See

THEOREM VIII., being an extension of the property of Theorem VI.-If a straight line be drawn to cut any two sides of a triangle, and the third side, one or all the sides being produced if necessary, thus dividing them into six segments (a prolonged side being taken as one segment, and its prolongation as another), then will the product of any three of these segments, which are not adjacent, be equal to the product of the other three.

to prove that AA′ × BB′ × CC'¬AC′ × BA' × CB'. Demonstration.-Through C, parallel to AB, draw CS, meeting

K

the straight line in which the points A', B', and C' are situated,

in S.

Then, by similar triangles C''AA' and C'C'S, we have AC': AA' : : CC': CS.

Also, by similar triangles A'B'B and CB'S, we have BA': BB':: CS: CB'.

By multiplying the corresponding terms, we have AC' × BA': AA' × BB':: CC': CB'; whence AA' × BB' × CC'

× CB'. Q. E. D.

AC' × BA'

THEOREM IX.-Conversely, if in the three sides of a triangle, or these sides produced, three points be taken such that the product of any three of the six segments not adjacent, into which the sides with their prolongations are divided, shall be equal to the product of the other three segments, then will the three points so taken be in the same straight line.

Demonstration.-We have to prove that if AA' × BB' × CC' : AC' × BA' × CB', then the three points A', B', and C' will be in the same straight line.

Parallel to AB draw CS to B' and C' are situated, in S.

meet the line in which the two points

Then, if this line does not cut AB in

A' (Fig. 1), suppose it to cut AB in some other point, as F. Then, by Theorem VIII., AF × BB' × CC' BF × CB' × AC",

But AF is greater than AA', wherefore BF is greater than BA', a part greater than the whole, which is absurd. Wherefore the proportion AF: AA': BF: BA' can be true only when F coincides with A'. Hence the three points A', B', and C' are in the same straight line. The same may be proved in like manner by either of the other figures.

NOTE. It will be observed in each of the figures under Theorems VI. and VIII. that each side has two segments; and bearing in mind that when a side is produced the whole line thus formed is one segment, and the produced part is the other segment, then taking the sides in order, and beginning at any angle (as A), take the segment (AA') terminating there of the first side; then the segment (BB′) of the second side, terminating at B; then the segment (CC') of the third side, terminating at C, and place their product equal to the product of the remaining three segments (BA′X CB' × AC′), and we have AA' X BB′ × CC' BA'X CB'X AC'. By a little attention in this manner, the order in which the segments are to be taken to form the equation in any case, may be readily fixed on the mind of the student.

THEOREM X.*—If a quadrilateral figure be in any manner divided into two quadrilateral figures, and diagonals be drawn to the whole. quadrilaterals, and also to the two partial ones, then the points of intersection of these three pairs of diagonals will be in the same straight line. Required, the demonstration.

Demonstration.-We have to prove that the points G, H, and I, which are the intersections of the three pairs of diagonals of the quadrilaterals ABED, ACFD, and BCFE, respectively, are in the same straight line.

CASE 1. When the lines DF and AC are paralleť (Fig. 1).

In similar triangles BIC and FIE, we have BI: FI::BC:EF; hence BIX EF = FI× BC.

In similar triangles FHD and CHA, we have FH: AH:÷DF:AC; hence FHX AC AHX DF.

* See Silliman's Journal of Science, vol. xxiii. p. 224, Second Series. See, also, Gillespie's Land Surveying, p. 387.

In similar triangles AGB and EGD, we have AG:EG::BG: DG.

By composition, we have AG: AE:: BG: BD; hence AG × BD = AE × BG.

Since triangle ABG is cut by the line CLD, we have, Theorem VIII., ALX BC × DG = AC × BD × GL.

Since triangle DEG is cut by the line AKF, we have, Theorem VIII., GK × DFX AE=DK × EF× AG.

Since triangle AGK is cut by the line HLD, we have, Theorem VIII., AH× GLX DK = AL × DG × HK.

By multiplying the terms on each side of these six equations together, and observing that the twelve terms EF, AC, AG, BD, AL, BC, DG, DF, AE, AH, GL, and DK on one side cancel the same quantities on the other side, we have BIX FHX GK=IF× BG × HK; whence, by Theorem IX., the points G, H, and I are in the same straight line, the triangle whose sides are cut being KBF.

CASE 2. When the sides DF and AC are not parallel (Fig. 2), let these sides be produced, and they will meet in some point, as P. Then, since triangle BFP is cut by the line CIE, we have, Theorem VIII., BI× EF× CP=BC × IF × EP.

Since triangle AFP is cut by the line CHD, we have, Theorem VIII., ACX FHX DP = AH× DF × CP.

Since triangle ABG is cut by the line CLD, we have, Theorem VIII., ALX.BC × DG = AC × BD × GL. Since triangle DEG is cut by the line BAP, we have, Theorem VIII., BDX EP × AG=DP × AE × BG.

Since triangle DEG is cut by the line AKF, we have, Theorem VIII., DFX AEX GKDKX EFX AG.

Since triangle AGK is cut by the line DLH, we have, Theorem VIII., AH X GLX DK= ALX DGX HK.

By multiplying the terms on each side of these six equations respectively together, and observing that the fifteen terms EF, CP,