[ 44« ] 
and like to the Arc A n H. Confequently AMC reprefenfs 
the Evolving Periferie, in a pofition like to the Arc A n H, 
and C is the defcribing point. 
6. After the fame manner may be found other points thro 
which the Curv may be drawn. But here ("as in the old Qua- 
</r^mx of the point E cannot be precifely deter- 
mined but the Curv may be brought fo near it, that its flexure 
or tendency will fo lead to the point E, that A E ftiall be 
near enough to the truth for common ufes. 
7. Suppofing the point E found, a Tangent to any point of 
the Curv may be drawn : and fuppofing a Tangent drawn,the 
point E may be determined,* the property of the Tangent 
being this, that fuppofing R T a Tangent to the point C and 
C A, CE, drawn from C to each end of the redlify'd Circle, 
the Angle ACT ( the leffer angle that AC makes with the tan- 
gent^ is equal to the tangent made by the 2Lines drawn fromC. 
8. c be a point in the Quadratrixindifinitely near to C • 
and draw A c interfedting AHKA in h, and AMC in o . To 
A c asa chord, draw the Arc Amc like unto the Arc A n a- 
To the point c of the Arc AMC draw the Targent CL 
= AE, and Joyn L A : So is oC an indefinitely little parti- 
cle of the Arc coincident with its Tangent. 
9. Eecaufe of the like Segments AnhA, AMoA, AmcA, as 
chord Ac: to chord lo : : So is Arc Amc (= AMC) ; to Arc 
AMo. Or Ac : Ao : Amc (= AMC; : AMo. And dividing 
Ac » Ao C= co) : Ao : : AMc—AMo (=Co; : AMo. That 
is, CO : Ao : : Co : AMo. and alternately, co : Co : : Ao : 
AMo. Put AC for Ao, and AMC for AMo (as difl^ering in- 
finitely little) and then 'tis co : Co ; : AE : AMC. But by 
conftrudion CL=AE =AMC whence co : Co . : AC : CL 
and the Angle LCA==Coc. ( oc being infinitely near to AC, 
is therefore parallel to it.J and therefore Coc, ACL are like 
Triangles, 
10. Becaufe of CL=AE, Ang. EAC = LCA. fCLand 
EA being Tangents to the two ends of the fame circular 
Arch AMC, make equal Angles with its chord AC.) and AC 
common to both, the Triangles E AC, ACL are like and equal i 
therefore are all three Coc/ ACL, EAC like Triangles. 
Whence it follows, that the Angle ACE fin the Triangle 
EAC) is equal to the Angle ocC (\n the Triangle coC.) 
But ocC = ACT becaufe oc and AC are parallel: there, 
fore Ang. ACE =ACT. Q£D. 
