262 
THE APPLICATION OF A LG ERRA 
PROBLEM XIII. 
The perimeter AB 4 BC-t-CA, and the perpendicular, 
BD,falling from the right angle B, to the hypothenuse 
AC, being given to determine the triangle. 
Let BD — a, AB zz x, BC zz y, AC =2, and AB 
4- BC + CA =r b : then, by reason of the similar tri 
angles ACB and ABD, it will be as z : y :: x : a; and 
q therefore xy zz az: more 
over, x 2 f- if — z z (by Euc. 
4 7- 1.) and x + y + z 
— b (by tlie question). Trans 
pose z in the last equation, 
arid square both sides, and 
you will have x 2 f 2xy 4- 
y z — b 2 — 2bz 4- from 
which take x 1 4- y % zz z x , 
arid there will remain <2xy — b 7 — <ibz\ but, by the 
first equation, 2xy is rz 2az; therefore 2az zz b l — 2bz 
b % 
and 2 rr — ; whence z is known. But to fmd 
2 a 4- 2 b 
x and y from hence, put ^--;g = c, and let this value 
of 2 be substituted in the two foregoing equations, 
x 4- y — b — z, and zy zz az, and they will become 
x 4- y zz b — c, and xy zz ac: from the square of the 
former of which subtract the quadruple of the latter, 
so shall x* — Qxy 4- yf — b — 4ac ; and conse 
quently x — y zz Vb — ci a — 4ac. This equation be 
ing added to, and subtracted from x 4- y zz b — y> 
gives 2x zz b — c 4- v/b — c Y — 4ac, and 2y zz b —. 
c — >Z b — c 1‘ — 4ac. 
PROBLEM XIV. 
Having the perimeter o f a right-angled triangle ABC, 
and the radius DF, of its inscribed circle; to determine 
all the sides of the triangle,