TO GEOMETRICAL PROBLEMS. 
i . t.t\ i D x Ai) b x x. a I ii i 
angle ABD being 0 r — we shall have, 
0 2 2 
b — x x ax* b — x x a ,, . 
—— : [by the question)-, and 
x x ax 
2 x c 4 •—• bx 
. . _ 7)lJ)X 711C 1 
consequently, nx 4 m X c z --bx, oi\r 2 -f — r: ; 
The geometrical construction of this problem, from 
the equation ¿r 2 -f —— r= niay be as follows. In 
CB let there be taken, CII : CB :: m : n, and let HK 
be drawn parallel to BA; then CHbeing = and CK 
mC _ Ml 1 1 . r>u 
— —, our equation will be changed to x 4 4- x x CH 
= AC x CK, or to CD x CD + CH - AC x CK. 
Upon CH as a diameter let the circle CTHQ be describ 
ed, in which inscribe CG — AK; and in CG produced, 
take CS — CA; and from S, through the centre O, 
draw the right line STOQ, cutting the circumference 
in T and Q, and make CD — ST; then will D be the 
point required. For CG being = AK, and CS = CA ; 
therefore will AC x CK =r CS x C'S “ ST x SQ (Euc. 
37. 3.) = ST x ST + TQ = CD x CD + CH, the 
very same as above. 
PROBLEM XXI. 
Having the perimeter of a right-angled triangle. ABC, 
and three perpendiculars DE, DF, and DG, falling 
from, a point within the triangle upon the three sides 
thereof; to determine the sides. 
Suppose DA, DB, and DC to be drawn; and let DE 
- a, DF - b, DG = c, AB - s, BC = y, AC = z.