Syntaxis mathematica, or, A construction of the harder problemes of geometry with so much of the conicks as is therefore requisite and other more ordinary and usefull propositions inter-mixed, and tables to several purposes / by Tho. Gibson.

Probl. 2. Ʋpon a line given as a Base to describe an Isos∣celes triangle, so that an inward parallel Base may cut off two segments of the sides betwixt the bases, so that either segment may be equal to the inwad base, the perpendicular from the vertex to the said inward Base being also by supposition given.

Let there be given the line cd, and the line bg.

And let it be required upon cd, as a base, to describe the Isosceles bcd, so as the line bg, fal∣ling at right angles with fe, equidistant to cd, the lines df, ef, and ec, may be equall each to other.

Put gr=e, and fe=a, And cd=b and bg=d, Then d′, d + e″, a′ b″,

[illustration] mathematical diagram

That is da + ea=db. Euclid. 6.16. which is the same aequation as was first in the former Probleme; and therefore if there be in the roome of da + ea=db substituted another equation like that in the former, such is the aequation.

a ⁴ − 2daaa + ffaa + 2 dcca − ddcc=o.

And that purged from the second term, as be∣fore, there ariseth a second equation.

+ eeee − mee + ne − p=o

And lastly, if d be a parameter, according to which a Parabola is described, the root e, and consequently the root a, may be found as in the former.

And thus having shewed the method general for all Aequations which attain but 3 or 4 dimensi∣ons, and exemplified it by Problemes which lead to such aequations; I now say that was the end of my present businesse. And if any still desire a lon∣ger reach, I referre him to Des Cartes, who hath proceeded to aequations of 5. and 6. dimensions; by which foure Mean proportionals, and quin∣quisection of Angles, and other sursolid Pro∣blemes may be found and effected.

Note 1.

The first Probleme of this Chapter, as it is more composed then trisection, so it comprehends it; as may be seen by (not only Pappus and others who applied it to that end but) the following Example. In which, let there be an arch of a cir∣cle, namely bc, and let it be required to divide the arch bc, into three equal parts, or (which is as good) to find the third part of the arch bc.

Find out the center e, and describe the circle, and draw the Diameter bg, and produce it to p, or further, as is need, and make cr the right sine

[illustration] mathematical diagram

of the arch bc, and from the center e; draw ex parallel to cr, and complete the rectangle cxer, and by the first Probleme, draw cp to cut ex, in f, so as fp, may be equal to bg, then I say that g•• is equal to a third part of bc.

From o through e, draw oez.

Now because eo is Radius, fp the diameter,

and the angle fepa right angle, therefore the lines fo, po, eo, are severally equal.

And the angles zeb=geo. and peo=epo.

And also foe=2peo. Euclid. 1.32.

Therefore also foe=2zeb.

But foe, that is coz, being in the peripherie, is measured by half the arch cz, Euclid. 3.20. wherefore bz which is the measure of half the angle coz, is a fourth part of the whole arch cz; and consequently a third part of bc, and therefore go which is equal to bz, is also equal to a third of bc, which was to be proved.

NOTE 2. Mechanically.

Seeing in this Scheme the line fo, is ever equal to the Semidiameter eg, if in commune practise there is bc, or any other arch (not greater then a quadrant) to be trisected, laying a thin Ruler to touch the point c, and cut the diameter bg, produced in p, the point required, the compasse being opened to the just length of Radius eg, setting one foot in ex, and shifting the Ruler till the other foot fall in the peripherie at o, the point o, shall always be distant from g, by a third of bc, the doing of which (although it must not be called Mathematical, yet) is very neer as easie, and as free from erring as from a point given to a point given to draw a streight line; or upon a center given with a distance given to describe a circle; & from a given point in it to set off an arch

equal to an arch given: And therefore I recom∣mend this as the most simple and short and safe way for Mechanick use.

NOTE. 3

If the arch to be trisected be greater then a quadrant, then trisect the complement thereof to a semicircle; and the third of this complement taken from 60 degrees (which is always a given arch) leaves the third of thé arch required.

Inscription of Chords in a Circle, and making aequicrurall triangles whose angle at the base shall be to the angle at the vertex in any given proportion, are the same thing: for to inscribe a figure of 3, 4, 5, 6, 7, 8, 9, 10, 11, sides &c, findes such triangles whose said angles shall be as 2/2, 3/2, 4/2, 5/2, 6/2, 7/2, 8/2, 9/2, 10/2, &c, as is easie to be seen by the operation.

Quadrature of the Circle is that in which (as yet) onely Archimedes hath laboured with any successe; he having demonstrated that the Cir∣cumference is to the Diameter lesser then as 22, to 7, and greater then 21 70/71, to 7, within which strict limits, a French man many years since found it to be in whole Numbers. thus,

As the whole circumference, is to the perime∣ter of the inscribed Square: so is 10, to 9, that is. Quadrant′. Subtense″. 10′.9″. which is easie from practise; and may be proved. thus,

Put the Diameter=7, the halfe=3½, of

which the squar is=49/4, and doubled is 49/2, whose square root is the side of the inscribed square. The whole perimeter therefore is 4√49/2, and the whole circumference is found by this Ana∣logisme.

9′.10″.4√49/2′.40/9√49/2″.

It rests to be proved that 40/9√49/2 is greater then 21 70/71, and lesser then 22.

Now 4,949, is somthing lesse then √49/2 which multiplied by 40, and divided by 9, the quotient it 21, 995, which yet is greater then 21 70/71. for 995/1000 > 70/71.

Again 4, 950. is somthing greater then √49/2, which multiplied by 40, and divided by 9, as be∣fore, the quotient is 22, so that 40/9√49/2 < 22. and 40/9√49/2 > 21 70/71. which was to be proved