( 'S6z ) 
Lines ax xh '^ whofe Squares together (hall be ctjual 
to the Square given gg. 
Le axb whofe height is xy be the Triangle required 
hika ab in and draw twx, 
AMALrSlS. 
Let therefore axa xbx = gg 
But By the ij^^^ofth^ Imtod.iijc^ -i-xbx = 2ama-+2mxni 
Therefore gg = aama 2 mxm 
or gg — 2ama = 2mxni 
Therefore the Problem is folv'd , but the Length of mx 
being given and not its Pofition , it is evident that it 
may be the Semidiameter o^ a Circle whofe Circumfe- 
rence fliall be the Locus of the point x. 
ConfiruBion And Demonftration. 
From the Square given gg Subtraft the double Square 
of am^ the Square root of half the remainder fliall be 
the line mx y with the Center «1 and diftance mx^ de- 
fcribe the Circle fxd^ I fay that any point x taken in its 
Circumference resolves the Problem, 
For fince the double of the Squares of am and xm is 
equal to the Square gg^ by the Conftruftion , and by the 
iph. Propofition of the Introdu'flion to the Squares 
and xb : The two Squares ax and xb together will be 
equal to the Square gg. Which was to be done. 
' FINIS. 
ERRATA. ~" 
"|3\ge3 5? l I. for IV. r. III. p. 356. 1. 26. for III. r. IV. and for /«^^ 
L^-.jff-'if^^'^^^^'^'h ^-^c. r.Juhftra^, 6cc. p. 357« - 33- Sojtgenes. 
