PROB. XIX. How to divide a Triangle, whose area or content is known, into two parts, by a line drawn from a point limited in any side thereof, according to any number of Acres, Roods and Perches.
IN the foregoing Triangle ABC, whose area or content is 5 Acres, 1 Rood, let the limited point be E in the base thereof, and let it be required from the point E to draw a right line which shall divide the Triangle into two parts between M and N, so that M may have 3 Acres, 3 Roods thereof, and N may have 1 Acre and 2 Roods thereof.
First, reduce the quantity of N (being the lesser) into perches, which makes 240, then (considering on which side of the limited point E this part is to be laid, as towards B) measure that part of the base from E to B 30 Perches, whereof take the halfe, which is 15, and thereby divide 240, the part belonging to N; the quotient will be 16, the length of the perpendicular FH, at which parallel distance from the base BC cut the side AB in F, from whence draw the line FE which shall cut off the Triangle FBE, containing 1 Acre, 2 Roods, the part belonging to N, then will the Trapezia AFEC (which is the part belonging to M) contain the residue, namely, 3 Acres, 3 Roods.