CEO METRICAL PROBLEMS. 
377 
FR ( = CH) whence AF will be given ; and then, AD 
and AF being given, AE —; %/AD x. AF will also be 
given. 
PROBLEM lx m. 
To find a point in a given triangle ABC, from whence 
right lines drawn to the three angular points, shall divide 
the whole triangle into parts, \ COA, AOB, BOC,) 
having the same ratio one to another, as three given right- 
lines, m, n, and p, respectively. 
CONSTRUCTION. 
In CA and AB produced, if need be, take CF, arid 
AF, each equal to m + n + p, joining E, B and F, C; 
take Ce — in, 
Ac — n, and 
draw eh and cf, 
parallel to EB 
and CF, meet 
ing the sides of 
the given tri 
angle in h and 
f\ draw also 
¿Q and /P pa 
rallel to AC 
and AB, and 
at O, the in 
tersection of these lines, will be the point required. 
DEMONSTRATION. 
Let ¿H and BD be perpendicular to AC. The tri 
angles CBE, Che, as also GBD, C6FI are similar; 
therefore, m (Ce) : m n F p ( CE) : : Ch : CB : ; 
¿>EI : BD : : the triangle AOC : triangle ABC. In 
the very same manner it mayEe proved, that the part 
AOB is to the whole triangle ABC, as n to m -f n 
(- p ; whence it follows, that the remaining part BOC 
must he to the whole triangle, as p to m +' n F p ; 
therefore these parts are to one another in the given 
ratio of m, n, and p. Q. E. I).