C C 2 
GEOMfcTRtCAL PROBLEMS* 387 
dicular thereto, and equal to half the given bisecting 
line: and from G, as a 
centre, with the radius 
GB, let a circle BIIF 
be described, intersect 
ing EG (when drawn) 
in F and H; from E A 
to A B draw ED — FF, 
and let the same be pro 
duced to meet the eir- 
cumference in G ; join 
A, C, and B, C ; so shall ABC be the triangle re 
quired. 
DEMONSTRATION. 
The triangles CBE and BDE are similar, because 
the angle BEC is common to both, and the angles BCE 
and DBE stand upon equal arches BE and AE: there 
fore EC : EB :: EB : ED, and consequently ED x EC 
= EB 1 : but [by Euc. 36. 3.) EB 2 = EF x EH - ED 
x EH [by construction). Hence ED x EC zr ED x 
EH, and consequently EC n EH; from which taking 
away the equal quantities ED and EF, there remains 
DC — FH — the given line bisecting the vertical angle 
(by construction) : and it is evident that DC bisects the 
angle ACB, since ACD and BCD stand upon equal 
arches AE and EB. Q. E. D. 
METHOD OF CALCULATION. 
If BE be considered as a radius, BR (f A.B) will be 
the co-sine of the angle EBR, and BG the ngent of 
BEG; therefore BR : BG (or AB : DC) :: co-sin. EBR 
(ACE) : tang. BEG, whose half complement FHBis 
likewise given from hence: then, the angle HBA (sup 
posing EB produced to6) being the complement of EHB, 
we shall have tang. EHB : rad. (.*: sin. EHB : co-sin, 
EHB :: BE : EH : EB : : EC) : ; sin. ECB : sin. CBE 
z= sin. EDB zr co-sin. OED, half the difference of the 
angles (ABC and BAC) at the base.