Lawson's Geometrical Theorems. ‘451 
Upon HE produced as a diameter de- 
scribe a semi-circle through H and A 
meeting it in K, then EK: AE = GE: : 
AE=GE:HE::HE: FE; take ME 
‘= EK, and parallel thereto NG = HE, 
and OE = NG; then NG: ME: : FE: 
GE, and by division NG: MO: : FE: 
FG, and MO: FG::NG=HE: FE:-: 
GE ~ NO :HE=NG, or MO: NO:: 
FG: NG, therefore the triangles FGN, 
MON are similar, and the angle NMO= 
NFE, consequently the points E, M, N, 
Fareinacircle. And conversely when 
these points are in a circle, and FE : 
NG: NG: GH, take EO = NG, and the 
triangles NOM, NGF are similar; there- 
fore F@ :OM:: NG: NO=GE :: by 
hypothesis FE : NG, and NG — OE: 
OM :: FE: FG, and by division NG: 
ME:: FE: GE, or FE: NG:: GE: 
ME, consequently FE : NG: : NG: 
triangle ABC, and if on BA or BC from B be put a third 
proportional to the other side and the bisecting line; then 
Isay the rectangle under that side on which it is put and 
its remainder when the third proportional is taken from it 
is equal to the square of the adjacent segment of the base 
made by the bisecting line. i. e. BCE==CD?, or BAE 
"== AB?, 
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