384 THE CONSTRUCTION OP 
KI, by the last problem, parallel to RG, so as to form 
the triangle KGi equal to the square G'S'i'V, and the 
thing is done. 
DEMONSTRATION. 
Since, by construction, KGI ( — GSTV) = QN, 
let AFEG — OQ be taken away, and there will re 
main AFE1K r; LN. Moreover, since the angle HGI 
is« — HGK, and the angle IMG (1IGR) a right one, 
the angles I and K are equal; and therefore, by the 
preceding lemma, iK is the shortest right-line that can 
possibly be drawn to cut otF the same area. Q. E. D. 
METHOD OF CALCULATION. 
Let the area of the figure AFFG be found, by di 
viding it into triangles AFG, EFG, and let this area 
be added to the given area to lie cut'off; and then, the 
square root of the sum being extracted, you will have 
GS the side of the square GT; from whence GI will be 
determined, as in the last problem. 
Note. In the same manner may a given area be cut 
off, by a right-line making any given angles with the 
opposite sides. 
PROliLEM LXX. 
Through a given point P, to draw a right-line FED to 
cut two right-fines AB, AC given b;/ position, so that the 
triangle ADE,' formed from thence, may be of a given 
magnitude. 
CONSTRUCTION. 
Draw PFH parallel to AB, intersection AC in F; 
E/V