to 2- Mr, hutton’s Demonstration 
added), the z _ lef = u + (3 + y- $+s, Sec. All this is evi- 
dent from Eucl. I. 29.; the angles we fpeak of being 
thofe, which are meafufed by the little arcs defcribed 
about each angular point in the figure. 
I. If right lines, ah, bi, ck, dl, em, be drawn from 
the angular points, perpendicular each to the parallel 
which pafles through the next angular point, the fums of 
the perpendiculars, drawn in contrary directions, the one 
upwards, and the other downwards, will be equal. And 
each perpendicular will be a fourth in proportion with the 
radius, that fide of the polygon which is adj acent both to the 
perpendicular and the parallel on which it falls, and the 
fine of the fum of the external angles taken to that inclu- 
fively from which the perpendicular is drawn. Thus, 
ah + bi + DL — CK + em ; and rad. fin. a. - ab : ah; and rad. 
fin. cc + jb = b c : bi, and in like manner of the reft. 
Take the value, therefore, of each perpendicular by 
thefe analogies, putting unity for the radius, fubtraCfc 
the fum of all that are drawn upwards from the fum 
■of all that are drawn downwards, and the remain- 
der, put equal to o, is the firft equation ; that is, 
ah - ax f.oc (for jL abh = a); bi = b x f. a. + (3 (for 
f. z. icb = f. /_ hbc = f. of its fuppl. obc or oc + /?) ; in like 
manner, ck = - c x f. a. + /3 + y; dl = d x i\ a + (3 + y — i; 
em = — exl.a + (3 + y— d' + 5, See. the laft perpendicular 
will always be = o, becaufe the fine of 360°, or of 
a+f3+y~S+s + £ is nothing. Hence a x f. ot + b x f. 
a. + /3 + c x {. ct + (3 + y + dx a. + f3 + y — $ + e x i. 
a + + y - c) N + s - ah + bi - ck + dl - em = o, which is 
the lirfik equation. 11. 
