BAG =r ACE (Euc. 32. l.) rr 
Q. E. D. 
METHOD OF CALCULATION. 
In the triangle ADE are given the sides AD, AE, and 
the angle D, whence the angle A will be given ; then 
in the triangle ACE are given all the angles and the side 
AE, whence AC andCB (CE) will be given likewise. 
PROBLEM XVII. 
The difference of the angles at the base, the ratio of 
the segments of the base, and cither the sum of the sides, 
the difference of the sides, or the perpendicular, being 
given, to construct the triangle. 
CONSTRUCTION. 
Let AC be to BC in the given ratio of the segments 
of the base; and upon AB let a segment of a circle 
BPA be described [by Problem 4.) to contain an angle 
equal to the difference 
of the angles at the 
base; raise CP per 
pendicular to AC, cut 
ting the periphery of 
the circle in P; and in 
AC produced, take CD 
= CB, and draw PA, 
PB and PD: then, if 
the perpendicular be 
given, take PF equal 
thereto, and, through F, draw EFG parallel to AD; 
' 'but if the sum or difference of the sides be given, let a 
fourth proportional PE, to AP + PD, AP, and the said 
sum or difference be taken, and draw EFG as above; 
then will PEG be the triangle required, 
DEMONSTRATION. 
Since CP is perpendicular to AD, and CD= CB, the 
angle D will be equal to DBP — A BPA: whence, 
because EG, is parallel to AD, PGIa will be — PEG 
i BPA [Euc, 2y. l.) and consequently PGE — PEG 
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