Geometry Simplified

If the method of tracing the line ACN by the movement of a parabola seems inconvenient to you, it is easy to find several other ways to describe it.

For example, if having the same quantities as before for AB and BL and the same for BK, which was taken as the principal latus rectum of the parabola; we describe the semicircle KST whose center is taken arbitrarily on the line BK, such that it intersects the line AB somewhere, as at point S, and that from point T, where it ends, we take towards K the line TV, equal to BL; then having drawn the line SV, we draw another parallel to it through point A, such as AC;

We also draw another through S, parallel to BK, such as SC; the point C, where these two parallels meet, will be one of those of the sought curve. And we can find, in the same way, as many others as we desire.

Or la demonstration de tout cecy est assés facile. car appliquant la reigle AE auec la Parabole FD sur le point C ; comme il est certain qu’elles peuuent y estre appliquées ensemble, puisque ce point C est en la courbe ACN, qui est descrite par leur intersection ; si CG se nomme y, GD sera yy/n, à cause que le costé droit, qui est n, est à CG, comme CG à GD. Et ostant DE, qui est … de GD, on a …, pour GE.

Puis à cause que AB est à BE, comme CG est à GE ; AB estant 1/2 p , BE est …

Et tout de mesme en supposant que le point C de la courbe a esté trouué par l’intersectiõ des lignes droites, SC parallele à BK, et AC parallele SV. SB qui est esgale à CG, est y: et BK estant esgale au costé droit de la Parabole, que iay nommé n, BT est yy /n. car comme KB est à BS, ainsi BS est à BT.

Et TV estant la mesme que BL, c’est à dire …, BV est … :

et comme SB est à BV, ainsi AB est à BE, qui est par consequent

…

comme devant, d’où on voit que c’est vne mesme ligne courbe qui se descrit en ces deux façons.

Aprés cela, pource que BL et DE sont esgales, DL et BE le sont aussy : de façon qu’adioustãt LH, qui est .., à DL, qui est .., on a la toute DH, qui est

…

et en ostant GD, qui est .. on a GH, qui est

…

ce que i’escris par ordre en cete sorte

GH =

Et le quarré de GH est,

….

Et en quelque autre endroit de cete ligne courbe qu’on veuille imaginer le point C, comme vers N, ou vers Q, on trouuera tousiours que le quarré de la ligne droite, qui est entre le point H et celuy où tombe la perpendiculaire du point C sur BH, peut estre exprimé en ces mesmes termes, et auec les mesmes signes + et -.

Moreover, since IH is m/nn and LH is …, IL is .., because of the right angle IHL ; and LP is ..

…

IP or IC is …, also because of the right angle IPL. Then, having made CM perpendicular to IH, IM is the difference between IH and HM or CG, that is to say between $m/nn$ and $y$, such that its square is always …, which being subtracted from the square of IC (Maire), there remains … for the square of CM, which is equal to the square of GH already found. Or else, by making this sum divided like the other by $nnyy$, we have … Then substituting …, for $nny^4$ and … for $my^3$: and multiplying both sums by $nnyy$, we have … equal to ….

That is to say that we have, … From which it appears that the lines CG, NR, QO, and the like are the roots of this Equation, which is what had to be demonstrated."

Thus, if we want to find four mean proportionals between the lines a and b, having set x for the first, the equation is x …

And doing … becomes ….

This is why we must take 3a for line AB, and … for BK or the right side of the Parable Mayor, p. 412 that I named n.

And … for DE or BL.

And after having described the curve ACN based on the measure of these three, we must make LH … and HI …

And LP … .

Because the circle which, having its center at point I, will pass through point P thus found, will cut the curve at the two points C and N; from which having drawn the perpendiculars NR and CG, if the smaller one, NR, is removed from the larger one, CG, the remainder will be, x, the first of the four proportional means sought.

It is easy in the same way to divide an angle into five equal parts, and to inscribe a figure of eleven or thirteen equal sides in a circle, and to find an infinite number of other examples of this rule.

However, it is to be noted that in several of these examples, it can happen that the circle cuts the parabola of the second kind so obliquely that the point of their intersection is difficult to recognize; and thus AT VI, 485 that this construction is not convenient for practice. To which it would be easy to remedy by composing other rules, in imitation of this one, as one can compose thousands of kinds.

I have reduced to a single construction all the problems of the same kind.

• But this also reduces them to an infinite number of other diverse ones, solving each of them in an infinite number of ways.

Example of these solutions are:

• constructing all those that are planar by intersecting a straight line with a circle
• constructing all those that are solid by also intersecting a parabola with a circle
• constructing all those that are complex by intersecting in the same way a line that is only one degree more complex than the parabola with a circle

We only need to follow the same path to construct all those that are infinitely more complex.

In mathematical progressions, when one has the first 2 or 3 terms, it is easy to find the others.

I hope that our descendants will be grateful to me for:

• the things I have explained here
• my deliberate omissions so that they can invent them

END