320 
TIIE CONSTRUCTION OF 
CONSTRUCTION. 
Draw the right-line AC at pleasure, in which take 
AB equal to the difference of the segments of the base, 
and make the angle CBE equal to half the supplement 
of the given angle ; 
and from A to BE 
apply AE equal to 
the given sum of the 
sides; make the an- 
. gle EBD = BED 
= and let BD meet A E 
in D, and from- the 
centre D, with the 
radius DB, describe 
the circle DBC, 
cutting AC in C, and join D, C; then will ACD be 
the triangle required. 
DEMONSTRATION. 
The angle EBD being = BED, therefore is DE = 
DB — DC, and consequently AD 4- DC =: AE. 
Moreover the angle CDE, at the centre, is double to 
the angle CBE, at the periphery, both standing upon 
the same arch CE; which last (by construction) is equal 
to half the supplement of the given angle, therefore 
CDE is equal to the whole supplement, and consequent 
ly ADC equal to the given angle itself. Q. E. D. 
METHOD OF CALCULTION. 
In the triangle ABE, are given tjie two sides AB, AE, 
and the angle ABE, whence the angle A will be given; 
then in the triangle ABD will be given all the angles 
and the side AB, whence AD and DC (DB) will be also 
given. 
PROBLEM VII. 
The angle at the vertex, the sum o f the including sides, 
and the ratio of the segments of the base being given ; to 
determine the triangle. 
CONSTRUCTION. 
Let AG be to GB, in the given ratio of the segments 
of the base, and, upon the right-line AB, let a segment