374 
THE CONSTRUCTION OF 
we have EHF - OAB + OB A (Euc. 32. l.) = 2OAB 
zz the given angle [by construction). Moreover, QH is 
— PK =: the given perpendicular; and EH being — 
AE, and FB - BP (Euc. 6. 1.), EH 1- HP + 
EF will therefore be = AB = the given sum of the 
sides. Q. E. D. 
METHOD OF CALCULATION. 
In the triangle AOP are given all the angles and the 
side AP, whence OP and AO (HO) are known; then in 
the triangle OHK will be given the »ides OH and OK 
(OP -f PK) whence HK will be given; next, in the 
triangle BQH will be given QH and BQ (BP — HK) 
whence QBH, and its double QFH, will be given; last 
ly, in the triangle EPH are given al l the angles and the 
perpendicular QH, whence the side^ will also be given. 
But the answer may be more easily brought out, by 
first finding, HOK, the difference of the angles ABH 
and BAH, as in the first Problem. 
problem lx 1. 
The sum of the three sides, the difference of the angles 
at the base, and the length of the line bisecting the x'ertical 
angle of any plane triangle being given; to describe the 
triangle. 
CONSTRUCTION. 
Make AB equal to the sum of the sides, which bisect 
in E by the perpendicular DErc, and make the angle n'E.r 
equal to half the 
given difference 
of the angles at 
the base, taking 
Er equal to the 
line bisecting the 
vertical angle; 
through r draw 
Cnr parallel to 
AB,cuttingDE« 
in * ; draw nA, 
to which draw Zlm ~ Er, and draw AD parallel to Em, 
meeting rcEDin D; and on the centre D, at the distance