Tq draic a right-line DF from one angle D of a given 
rhombus A BCD, so that the part thereof FG intercepted 
by one of the sides including the opposite angle and the 
ether side produced, may be of a given length. 
Let DE be perpendicular to AB; and let AB (= AD) 
= a, AE — b, FG 
rz c, and AF — x: 
then DF 1 ( = AF 1 
+ AD 2 -2AExAF) 
rr xx + aa — 2bx ; 
and, by similar tri 
angles, xx A- an — 
2fa'(DF I ): a\r(AF 2 ) 
:: cc (FG 1 ) : x — aV (BF 2 ); and consequently 
xx f aa—‘¿bx x xjT—Vaxf^aa zz eexx. Make ma — 
b, and na — c; so shall our equation become 
xx f aa—2niax x xx—Tax + aa ~ n'a'x': which, di 
vided by a'x 2 , gives— -g- — 2tn x + — 2 — 
J b a x ax 
?г 2 : this, by making z zz --- F —, becomes z—2m x 
2 zz n*: therefore — L 2m + 2 X z — n 1 — 4w, 
and z zz l 4- m + n* 4- 1 - 
a \ b 1- y/c y a by restoring the values 
m and n. From whence the value of x will be also 
iKq 
AC i