[412] 
This Semi-circle^ will he Equal to that S^adrant, ( Becaufe the 
Squares of their Diameters, are as a to i And, in (uch propor- 
tion are their refpedive Circles ; and therefore a Quarter or the 
one, equal to Half the other. ) 
And, conlequently, If, from each of thefe, we fabtra6l the 
common Segment ABD; iht Remain'mg Lunula ADBE i on the 
one fide ) will be Equal to the Remaimng Triangle ( on tlie other 
fide ) ABC. ( Or, to ABK, fuppofing AB bife^ted in K ; that is, 
to half the Square CK, mfcribed in the Leder Circle. ) Which 
IS commonly called, The Squaring of Hippocrates s Lunula ; That 
is, the Finding a ReBilinear Figure { which may be eafyly redu- 
ced to a Square ) equal to that Lunula, 
This being premifed ; The Point in hand, is, the Squaring a 
given Tortion of fuch Lunula : fuppofe ADE , cutt-ofi" by a 
Streight Line CDE, drawn from the Center C. Which Mr Terhs 
( not knowing that the like had been before attempted by any 
other ) doth perform after this manner ; viz. 
Drawing the Streight Lines E A, and EB ( cuttmgihe Arc EB 
in G, ) and, on AG, a perpendicular EF^ ( which will therefore 
pafs to the Center C, becaufe Bifeding AG at Right-angles ; ) The 
Right-lined Triangle AFE, is equal to ADE, the propofed Tort ton 
of the Lunula, 
His Demonflration is to this purpofe : viz, 
ADB being a Quadrantal Arc; the Angle AGB will be Thrh 
Hahes of a Right Angle ; ( and its Conjunit Angle EGA, Half 
a Right Angle. ) And that Angle ( being External to the Trian- 
gle AGE,) is Equal to the Two Oppofite Internals GEA + EAG. 
Whereof GEA ( becaufe an Angle in the Semicircle AEB) is a 
Right Angle; and therefore E AG is i%^a Right Angle, (as are 
alio PEG, and FEA. ) And the Three Triangles AFE, GFE, 
and 
