TRANSVERSALS. 
411 
27 
Fig. 418. 
of a triangle, lines le drawn, to a point situated either within or without 
the triangle, and prolonged to meet the sides of the tri 
angle, or their prolongations, thus dividing them into 
six parts, then will the product of any three non-con- 
secutive parts le equal to the product of the other three 
parts. 
That is, in Fig. 418, or Fig. 419, 
AE x BF xCD = EB xFOxDA. 
For, the triangle ABF, being cut by the transver 
sal E C, gives the relation (Theorem I). 
Fig. 419. 
AE x BO x FP = EB x FC x PA. 
The triangle A C F, being cut by 
the transversal D B, gives 
DCxFBxPA=ADxCBx 
FP. 
Multiplying these equations to 
gether, and suppressing the common 
factors PA, OB, and F P, we have 
AE x BF x CD = EBxFOx 
DA. 
Theorem IV. — Conversely : If 
three points are situated on the three 
sides of a triangle, or on their pro 
longations (either one, or three, of these points leing on the sides), so that 
they divide these lines in such a way that the product of any three non-con- 
secutive parts equals the product of the other three parts, then will lines drawn 
from these points to the opposite angles meet in the same point. 
This theorem can be demonstrated by a reductio ad absurdum. 
COROLLARIES OF THE PRECEDING THEOREMS. 
Corollary 1.— The MEDIANS of a triangle (i. e., the lines drawn from 
its summits to the middles of the opposite sides) meet in the same point. 
For, supposing, in Fig. 418, the points D, E, and F to be the middles of 
the sides, the products of the non-consecutive parts will be equal—i. e., 
AExBFxCD = DAxEB xFC; since AE = E B, B F = F C, C D 
= D A. Then Theorem IV applies. 
Cor. 2.—The BISSECTRICES of a triangle (i. e., the lines bisecting its 
angles) meet in the same point. 
For, in Fig. 418, supposing the lines A F, B D, CE to be bissectrices, we 
have (Legendre, IV, 17): 
BF:FC::AB:AC) (BF x AC = F C x AB, 
CD:DA::BC:BA whence CDxBA = DAxBC, 
AE : E B :: C A : C B ) ( AE x CB = EB x CA. 
Multiplying these equations together, and omitting the common factors, 
we have BF x CD x AE = FC x DA xEB. Then Theorem IV applies. 
Cor. 3.—The ALTITUDES of a triangle (i. e., the lines drawn from its 
summits perpendicular to the opposite sides) meet in the same point.