APPENDIX B. 
TRANS VERBALS. 
Theorem I.—If a straight line he drawn so as to cut any two sides of a 
triangle, and the third side prolonged, thus dividing them into six parts (the 
prolonged side and its prolongation being two of the parts), then will the 
product of any three of those parts, whose ex 
tremities are not contiguous, equal the product 
of the other three parts. 
That is, in Fig. 416, ABO being the triangle, 
and D F the transversal, BExAI)xCF = 
EAxDOxBF. 
To prove this, from B draw B G, parallel to 
0 A. From the similar triangles BEG and 
A E D, we have BG: BE :: AD : AE. From 
the similar triangles B F G and C F D, we have 
C D : C F :: B G : B F. Multiplying these proportions together, we have 
BG x OD: BE x OF :: AD xBG:AE xBF. Multiplying extremes 
and means, and suppressing the common factor B G, we have B E x A D x 
CF = EA x DC x BF. 
These six parts are sometimes said to be in involution. 
If the transversal passes entirely outside of the triangle and cuts the pro 
longations of all three sides, as in Fig. 417, the theorem still holds good. 
The same demonstration applies without any change.* 
Theorem II.—Conversely : If three 
points he tahen on two sides of a triangle, 
and on the third side prolonged, or on 
the prolongations of the three sides, di 
viding them into six parts, such that the 
product of three non-consecutive parts 
equals the product of the other three parts, 
then will these three points lie in the same 
straight line. 
This theorem is proved by a reductio 
ad absurdum. 
Theorem III.—If from the summits 
Fig.. 417. 
A 
Fig. 416. 
A 
* This theorem may be extended to polygons.