322 
TME CONSTRUCTION OF 
given angle; from A to Bn apply AE == the differ 
ence of the sides; pro 
duce AE, and make 
the angle EBO = 
BEO, and let BO meet 
AE, produced in O, 
and from the centre 
O, at the distance of 
OB, describe the cir 
cumference of a circle, 
cutting AB produced 
in C, join O, C; then 
DEMONSTRATION. 
Because the angle EBO is = BEO (by construction) ; 
therefore is EO = BO CO, and consequently AO 
— OC — AE. Furthermore because the angle AOC 
is double to ADC, and ADC = ABE (Eue. Corol.zz, 
3.) therefore is AOC also double to ABE. Q. E. D. 
METHOD OF CALCULATION. 
The two sides AB, AE, and the angle ABE being- 
given, the angle A will from thence be found ; then in 
the triangle ABO will be given all the angles and the 
side AB, whence OB (OC) and OA will be known. 
PROBLEM IX. 
The angle at the vertex, the difference o f the including 
sides, and the ratio o f the segments of the base, being 
given, to determine the triangle. 
CONSTRUCTION. 
cular to AB, meeting the 
Let AG be to GB, iu 
the given ratio of the seg 
ments of the base, and up 
on the right-line AB let a 
segment of a circle ACB 
be described (by Prob. 4.) 
capable of the given an 
gle ; draw GC perpendi- 
periphery in C, and join A, C