386 
THE CONSTRUCTION OF 
MN equal to AGHB, and, by the last problem, let ED 
be so drawn through the given point P, that the triangle 
A ED, formed from thence, may be equal to the whole 
rectangle KN; then will EDBHG be equal to KL ; 
for since AED is — KN, let the equal quantities 
AGHB and MN be taken away, and there will remain 
EDBHG - KL. 
METHOD OF CALCULATION. 
Let the area of the figure AGHB be found, by di 
viding it into triangles, and let this area be added to the 
area given, and the sum will be equal to the area AED, 
or the rectangle KN; from whence AD will be found, 
as in the last problem. 
PROBLEM LXXII. 
Having the base, the vertical angle, and the length of 
the line bisecting that angle and terminating in the base, 
to describe the triangle. 
CONSTRUCTION. 
Upon the given base AB let a segment of a circle 
ACB be described [by Problem 4.) to contain the give4 
angle, and, having completed- the whole circle, from 
0, the centre thereof, perpendicular to AB, let the ra 
dius OE he drawn; draw EB, and make BG perpen-