Errata

Lines $ax$, $xb$, whose Squares together shall be equal to the Square given $gg$. Let $axb$ whose height is $xy$ be the Triangle required. Bisect $ab$ in $m$ and draw $mx$. **Analysis.** Let therefore $axa + xbx = gg$ But by the 13th. of the Intro. $axa + xbx = 2ama + 2mxm$ Therefore $gg = 2ama + 2mxm$ or $gg - 2ama = 2mxm$ Therefore the Problem is solv'd, but the Length of $mx$ being given and not its Position, it is evident that it may be the Semidiameter of a Circle whose Circumference shall be the Locus of the point $x$. **Construction and Demonstration.** From the Square given $gg$ Subtract the double Square of $am$, the Square root of half the remainder shall be the line $mx$, with the Center $m$ and distance $mx$, describe the Circle $pxd$. I say that any point $x$ taken in its Circumference resolves the Problem. For since the double of the Squares of $am$ and $xm$ is equal to the Square $gg$, by the Construction, and by the 13th. Proposition of the Introduction to the Squares $ax$ and $xb$: The two Squares $ax$ and $xb$ together will be equal to the Square $gg$. Which was to be done. **FINIS.** **ERRATA.** Page 355. l. r. for IV. r. III. p. 356. l. 26. for III. r. IV. and for subtraction, subtraction, &c. r. subtraction, &c. p. 357. l. 33. r. Soligenes.