272 
TUE APPLICATION OF ALGEBRA 
of ACG) :: x(AC) : mx — AG; half whereof, mul- 
v l 
mcx — mx 
tiplied by c — x (BC), gives 
— a 1 , the area 
2 
of the triangle: from whence x comes out — \c + 
PROBLEM XIX. 
The hypothenuse AC, of a right-angled triangle A BC, 
and the side of the inscribed square BED!', being given; 
to determine the other tico sides of the triangle. 
Let DE, or DF n a, AC — b, AB — x, and BC r: y; 
, then it will be, as x: y :: x 
^ —a(AF) : a (I D); whence 
we have ax — yx — ya, and 
consequently xy — ax -f ay. 
E Moreover, xx 4- yy — bb: 
to which equation let the 
double of the former be add 
ed, and there arises x 1 4- 2xy, 
4- v 2 rr b 2 4- 2ax 4- 2an: 
X x + y, or x + y | a — 2a x o-’ + y — P'i where, by con 
sidering x 4- y as one quantity, and completing the 
square, we have x 4- y\ z — 2a x x+y -4- a z — b l 4- a z ; 
whence x j- y —a zz \Zb'~ 4- a 1 , and x y —\/d 1 4- b z 
4- a; which put = c: then by substituting, c—„r in 
stead of its equal (y) in the equation xy — ax 4- ay, 
there will arise cx — x 2 = ac ; whence x will be found 
= \c -f- \/?cc —~ac, and y — {c — \/\cc — ac. 
It appears from hence that c, or its equal \/ aa + bb 
+ a, cannot be less than 4a, and therefore b 2 not less 
than 8a 2 ; because the quantity ^cc — ac, under the 
radical sign, would be negative, and its square root im 
possible; it being known that all squares, whether from 
positive or negative roots, are positive; so that there 
cannot arise any such things as negative squares,