315 
THE CONSTRUCTION OP 
DEMONSTRATION. 
Because the angles BCD and CBD are equal, there 
fore is CD = DB (Euc. 6. 1.) and consequently AD -p 
DC — AB: likewise, for the same reason, the angle 
ADC — BCD f CBD, [Euc. 32. 1.) is equal to 2CBD. 
Q. E. D. 
METHOD OF CALCULATION. 
In the triangle ABC are given the two sides AB, AC, 
and the angle ABC, whence the angle A is known; 
then in the triangle ADC will be given all the angles, 
and the base AC ; whence the sides AD and DC will 
.also be known. 
PROBLEM II. 
The angle at the vertex, the base, and the difference of 
the sid-cs being given, to determine the triangle. 
CONSTRUCTION. 
Draw AC at pleasure, in which take AD equal to 
the difference of the sides, 
and make the angle CDB 
equal to the complement 
of half the given angle to 
a right angle; then from 
the point A draw AB 
equal to the given base, so 
as to meet DB in B, and 
make the angle DBC zz 
CDB, then will ABC be 
the triangle required. 
DEMONSTRATION.. 
Since [by construction) the angles CDB and DBC 
are equal, CB is equal to CD, and therefore CA — CB 
— AD: moreover, each of those equal angles being equal 
to the complement of half the given angle, their sum, 
which is the supplement of the angle C, must therefore 
be equal to two right angles — the (whole) given angle, 
and consequently C — the given angle. Q. E. D. 
METHOD OF CALCULATION. 
In the triangle ABD are given the sides AB, AD, 
C