[416] 
B H 2 d,) is equal to the Redlilineal Triangle AEF, or A s or 
" B 2 f, refpedlively. 
" And it fo happens, that, if this line going out from C, be on 
" the fame fide or the Diameter MN with the Lunula of Hippo- 
crates \ the forefaid Space (which receives a pcrfedt Quadra- 
" ture) is folitary ; ( fuch as are the Parts Hippocrates^ Lunula \ 
" and of the two Spaces A g CM, BHCN ;which therefore are Parts 
of the Lunula more nearly relating to one another.) 
" But if that Line going out from C, be on the other fide of 
"MN ; then the Space which is equal to the Re6lilineai Triangle, 
" IS, the difference of two Mixtilineal Figures, ( the one a Tri- 
"lineum, the other a Segment of the Ldier Circle,) as is above- 
"faid; neither of which can be fquared leverally. 
"All thefe particulars are plain from Mr. Terhs's Demonftra- 
" tion ; which, with a little variation ( fuch as is ufual in the dif- 
" ferent Cafes of the lame Theoreme) is applicable to all of them : 
" though perhaps he was not aware of it. 
" In the Dimenfion of the Parts of Hippocrates s Lunula^ 
" it might perhaps be expedted, that the Triangle afligned equal to 
" a Portion of the Lunula^ fliould be Part of the Triangle to 
" which that whole Lunula is wont to be afligned equal ; ( that is, 
" that the Triangle afligned equal to the Portion ADE, Ihould be 
the refpe6live part of ACB which is equal to the whole Lunula\ ) 
" which in that of Mr. Terks is not. 
"But, in that of Mr. Z/c/^/V^-^^w/d-C above-mentioned ) it is fo, 
which is to this pt>rpoie. 
" If from any Point E, in the circumference of the Lefler Circle, 
" we let fall on AB, a Perpendicular cutting it in L, and draw the 
" line CL ; the Triangle C AL, is equal to the Portion of the Lm- 
^mtla AED. ( And, confequently, the Tmngk CBI^ equal to 
"the Portion BED.) j . 
^y.^- Which ( becaufe Mr Tfchirnhaufe hath not at all done it ) 
" I fliall briefly Demonftrate, fo as the Deraonftration may reach 
" the Tortions of the Conjugate Space ACB g > A. 
" For the Triangles ACB, AEF, are like Triangles, each being 
"the half of a Square : And therefore, by ip el. <S, the Triangle 
" ACB IS to the Triangle AEF in the duplicate proportion of BA 
" to AE, that is, by 8, el. as BA is to AL. But, by i. el. <5, the 
" Triangle ACB is to the Triangle ACL as B A is to AL. There- 
" fore, by 9. el. the Triangles ACL and AEF are equal. But 
" the Triangle AEF is ( by Mr P<?r/^i" ) proved equal to the Por- 
" tion 
