C *44 ] 
let ^ Circle be dcfcribed, interfe&ing AG, produced 
m E, and the Line of Direction AT) in H> join 
, H , and Jet HI, AM and OR, be perpendicular 
to AE , AO, and ^P^refpedively, and let BC , pro- 
duced, meet AH in 2). 
It will appear, from what has been faid above, 
that AT ) 2 :TR 2 : : T>C : i? 5 therefore being 
= lAG 1 ^UE\ and RO=TO=±AE 
(by Conftruction), we ha vcAT) 2 :±AE: :T>C:±A£, 
and therefore AT) 2 =AExT)C. 
Now, the Triangles AT)C, AEH, being equian- 
gular (becaufe AT)C~T>AN ~ AEH, and TAG 
common to both) we likewife have AT) : T)C : : 
AE : EH, and confequently AE x T)C— ADxEH 
-AT 2 (per above) ; whence £#=^<D. There- 
iore, as the Triangles AT>B and EHI are equian- 
guiar, they are equal in all refpeds 5 and fo HI—AB: 
Y\ hence follows this eafy 
Conftruftion . 
Having defetibed the Circle AEF, as above di- 
icded, and drawn MG perpendicular to AE, take 
Co equal to AB, and thro’;?, parallel Mo AE, draw 
Hh, cutting the Circle in H and h 5 join A, H 
and A, hi then cither of the DiredionsTl# or Ah 
will anfwer the Conditions of the Problem. From 
this Conftrudion we have the following Calculation' 
As AB is to BC, fo is AG to OG; which added 
to, or fubtraded from, Gn {AB) gives On: Then 
it will be, as AG: On : : the Co-line of O AG ; Co- 
ftne of HOn {=.HAh) the Difference of the two 
required 
