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  "text": "Lines $ax$, $xb$, whose Squares together shall be equal to the Square given $gg$.\n\nLet $axb$ whose height is $xy$ be the Triangle required. Bisect $ab$ in $m$ and draw $mx$.\n\n**Analysis.**\n\nLet therefore $axa + xbx = gg$\n\nBut by the 13th. of the Intro. $axa + xbx = 2ama + 2mxm$\n\nTherefore $gg = 2ama + 2mxm$\n\nor $gg - 2ama = 2mxm$\n\nTherefore 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 $x$.\n\n**Construction and Demonstration.**\n\nFrom 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.\n\nFor 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.\n\n**FINIS.**\n\n**ERRATA.**\n\nPage 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.",
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