### The Quadrature of the Parabola, solved by many methods through the new geometry of indivisibles

##### Translation by Andrew Leahy and Kasandara Sullivan

*Until now, the matter about the measurement of the parabola has been related in the manner of the ancients. It remains that we should approach the same measurement of the parabola with a certain new but marvelous system--namely, by the aid of the Geometry of Indivisibles and with diverse methods in this manner. In fact, with the principal theorems of the ancients assumed, as much of Euclid as of Archimedes (they may concern very diverse matters themselves), it is amazing that from every single one of these the quadrature of the parabola is able to be elicited with so little trouble and vice versa, as if there were a certain common bond of truth. In fact, from the premise that a cylinder is three times its own inscribed cone, hence it follows that the parabola is 4/3 of its own inscribed triangle. Indeed, if you prefer to assume that the cylinder is 3/2 of its own inscribed sphere, the quadrature of the parabola is immediately inferred. The same result is concluded by supposing the demonstration which shows that the center of gravity of a cone is placed on its axis, so that the part which is near the vertex is three times the remainder. The parabola is no less squared by also supposing that the space bound by a spiral line in the first revolution and by the line which is the beginning of the revolution is 1/3 of the first circle. On the other hand, with the quadrature of the parabola supposed, all the aforementioned theorems are able to be demonstrated easily. Moreover, for my part I would not dare to assert that this Geometry of Indivisibles is a thoroughly new invention. Rather, I would have believed that the old geometers used this one method in the discovery of the most difficult theorems, although they would have produced another way more acceptable in their demonstrations, either for concealing the secret of the art or lest any opportunity for contradiction be proffered to envious detractors. Whatever it is, it is certain that this wonderful geometry is a shortcut for invention, and that it confirms countless almost inscrutable theorems with brief, direct, and affirming demonstrations, which is certainly not able to be done easily through the ancient teaching. To be sure, this is truly the Royal Road in the mathematical thorn hedges, that Cavalieri, creator of these wonderful inventions, first among everyone opened up and made public for the common good. *

#### Proposition 11.

The parabola is 4/3 of a triangle that has the same base and height.^{1}

Let ABC be a parabola with tangent CD and let AD be parallel to the diameter. Let the parallelogram AE be drawn, and let a circle with diameter AD be conceived which is the base of a cone having vertex at point C and likewise is the base of some cylinder ACED of the same height with the aforementioned cone.

Now let some line FG be drawn parallel to AD, and let a plane parallel to the circle on AD be conceived to pass through the line itself. FG will then be to IB as the line DA is to IB--that is, as the square on DC is to the square on CI (because of the parabola). Or as the square on DA is to the square on IG (because of similar triangles)--that is, as the circle on DA is to the circle on IG--namely, as the circle on FG is to the same circle on IG. And it is this always. All the first magnitudes are equal to the line DA and therefore equal among themselves. Also, all the thirds are equal to the circle on DA, and on account of this equal among themselves. Therefore, by Lemma 18, all the firsts together--namely the parallelogram AE--will be to all the seconds together--namely, to the trilineum ABCD--as all the thirds together--namely, the cylinder AE--are to all the fourths together together--that is, to the cone ACD.^{2} Therefore, the parallelogram AE is three times the trilineum ABCD.^{3} With the parallelogram AE cut in half, the triangle ACD will be 3/2 of the trilineum ABCD. By conversion of the ratio, the triangle ACD will be three times the parabola itself. On account of this, from the explanation of Proposition 9, the parabola will be 4/3 of its own inscribed triangle. *Quod erat &c.*

*We will also square the parabola by a different computation with the principles of indivisibles demonstrated previously, by which it will be able to be done with brevity. Moreover, we, scraping the earth with less daring, will resolve it by means of the immense ocean of Cavalieri's Geometry. Whoever wishes will be able to see all these things (shall I say in a fountain, or in a sea?) around the middle of the second book of Cavalieri’s **Geometry of Indivisibles*.

#### Lemma 20.

The squares of all the parts of a straight line together are in a ratio of 1/3 to the same number of squares of the whole line together.

*Let AB be a straight line. I say that all the squares of all the parts of the line AB together are 1/3 of the same number of the squares of that same line AB. *

*Indeed, let ACDB be a square with the diameter AD drawn. Let the figure be revolved around the axis AB until it returns to its starting point. It is clear that a cylinder CH is described by the square and also the cone DAH which has vertex at A is described by the triangle ABD. Now let EF be drawn parallel to CA and let AF--or FG (indeed they are equal)--be one of the infinite parts of the whole line AB. *

*Now the square on the whole line AB is to the square on the part AF as the square on EF is to square on FG (because of equality)--namely, as the circle on diameter EL is to the circle on diameter GI (Euclid XII.2). And it will always be so. Also, the first magnitudes individually are equal to the square on AB, and the thirds are always equal to the circle on DH. Therefore, by Lemma 18 all the firsts together (that is, as many lines AB squared as the line itself has parts) will be to all the squares of the parts as all the thirds together (that is, as the cylinder CH) will be to all fourths together (namely, to the cone DAH). Therefore, just as many squares of some line as the line itself has parts are to all squares of its very parts as the cylinder CH is to the cone DAH--namely, 3 to 1. By transposing, the proposition stands. **Quod demonstrandum fuerat &c.*

#### Lemma 21.

All rectangles which are bounded both by a straight line together with its own individual pieces and by the remainders are in a ratio of 2/3 to the same number of squares of the same straight lines.

*With the figure in the preceding lemma assumed, some point F should be taken on the line AB. The rectangle contained by BAF (as one straight line) and by FB will be one of all the aforementioned rectangles. (In other words, one side is composed from the whole line AB together with the part AF; the other side is clearly FB, without a doubt the remaining part.) *

*Moreover, the aforementioned rectangle contained by BAF as one line and by FB is the same as the rectangle EIL (because of the equality of sides). And this is always true in this manner wherever the point F is. But all the rectangles contained both by the lines intersecting in the trapezoid CAHD (one of which is EI) and by the remainders (one of which is IL) together with all the squares of the intermediate section (one of which is FI) are equal (because of Euclid II.5) to all the squares on the halves (one of which is FL). In fact, all squares of the intermediate sections (one of which is FI) are to all the squares of the halves (one of which is FL) as 1 is to 3 (by the preceding lemma). Therefore, if all squares of the intermediate sections are taken away, all the rectangles will remain (one of which is EIL). That is, all the rectangles contained both by AB together with its parts and by the remainders will remain as 2/3 of all the squares which are made from the halves--that is, the same number of all the squares of the whole AB. **Quod fuerat ostendendum &c.*

#### Proposition 12.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter is BE, and around the parabola let there be a parallelogram DC. Let a line FG be drawn parallel to the diameter. FG will be to GI as BE is to GI, or as the rectangle CEA is to CGA--that is, as the square on CE is to the rectangle CGA. And it will always be so. Also, the first magnitudes are always equal to the line BE. Moreover, the thirds are always equal to the square on CE. Therefore (by Lemma 18) all the firsts together (that is, the parallelogram AB) will be in relation to all the seconds together (namely, the semiparabola AIBE) as all the thirds together (clearly just as many squares on the line CE as the line itself has parts) in relation to all the fourths together (namely, all the rectangles contained by CE with its parts and by the remaining parts). Therefore (from the preceding lemma) the parallelogram AB will be 3/2 of its own semiparabola. Also, the whole parallelogram DC will be 3/2 of the whole parabola--indeed, as 6 is to 4. Therefore the parabola will be to its own inscribed triangle (which is in fact 1/2 of the parallelogram DC) as 4 is to 3--that is, 4/3. *Quod erat &c.*

*We are able to square the parabola by the same argument without the trouble of the former lemmas, with yet a different supposition--namely, by supposing the proportion which the cylinder has in relation to its inscribed sphere--a proportion which is indeed 3/2, as shown from Archimedes' **On the Sphere and the Cylinder, Book I*

#### Proposition 13.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola, around which is the parallelogram AD. Let there be a semicircle on the diameter AC, around which is the rectangle AE. Then with the axis AC fixed, let it be conceived that the semicircle itself is revolved around the axis in such a way that from its own revolution a sphere is circumscribed. A cylinder likewise arises from the revolution of the rectangle AE.

With some point G now assumed, let a line GF be drawn parallel to the diameter HB, and through the same point G let the plane GL be drawn perpendicular to the axis AC.

The line FG will be to GI as BH is to GI (because of equality). That is, as the rectangle CHA is to the rectangle CGA, or as the square on HN is to the square on GM (because of the circle), or as the square on GL is to the square on GM--namely, as the circle on the semidiameter GL in the cylinder is to the circle on the semidiameter GM in the sphere. And it will always be thus wherever the point G is assumed. Moreover, all the firsts are equal among themselves, as are all the third magnitudes among themselves. Therefore, all the firsts (namely, the parallelogram AD) will be to all the seconds (namely, to the parabola ABC) as all the thirds (that is, the cylinder) are to all the fourths together (clearly to the sphere). But the cylinder to the sphere is 3/2. Therefore, the parallelogram AD will certainly be 3/2 of the parabola, and the parabola itself will be 4/3 of its own inscribed triangle, as was concluded in the preceding lemma. *Quod &c.*

#### Lemma 22.

Suppose some number of magnitudes will have been hung from given points on a balance, and the same number of magnitudes of another type, equally proportional with the aforementioned magnitudes, will hang from the same points. The center of equilibrium of each type of the magnitudes will be one and the same.

*Let some number of magnitudes C, D, E and F of the first type be hung from whatever points on a balance AB. Let just as many magnitudes G, H, I, and L of a second type hang from the same points and let them be proportional--namely, as C is to D so is G to H, and as C is to E so is G to I, etc. I say that the same point on the balance is the common center of equilibrium of each type of suspended magnitudes. *

*Indeed, since as C is to D so is G to H, they will counterbalance from the same point, so the two magnitudes C and D will be as the two magnitudes G and H. *

*Further, since as C is to D so is G to H, by convertendo and componendo DC will be to C as HG is to G. Moreover, C is to E as G is to I. Therefore, CD together will be to E as GH together is to I by equality. Wherefore the magnitudes CD and E will counterbalance from the same point from which the two GH and I counterbalance. *

*Additionally, since as CD is to E so is GH to I by the things just mentioned, by componendo CDE will be to E as GHI is to I. But E is to C as I is to G, and C is to F as G to L. Wherefore CDE together will be to F as GHI together is to L by equality. Therefore, the two magnitudes CDE and F will have the same point of equilibrium which the two magnitudes GHI and L have. And thus also if there are many magnitudes, all the way to infinity, **quod erat propositum &c.*

#### Lemma 23.

If a parabola should have a tangent at the base and indeed a line parallel to the diameter for the other side, the trilineum bound by the parabolic curve, the tangent, and the aforementioned parallel will balance at equilibrium from the point on the tangent where it is divided so that the part nearest to the point of tangency is three times the remaining part.

*Let ABC be a parabola whose tangent at the base is CD and let AD be parallel to the diameter. I say that the mixed trilineum ABCD balances at equilibrium from the point on the tangent CD where it is divided so that the part towards the tangent C is three times the remaining part. *

*Let the figure be imagined in such a way that DA is perpendicular to the horizontal, and let a circle on the diameter DA be conceived which is the base of a cone having vertex at point C. *

*With some point E already assumed, let EF be drawn parallel to DA itself, and let a plane parallel to the base of the cone pass through the EF. *

*Therefore the line DA will be to EB as the square on DC is to the square on CE (because it is a parabola), or as the square on DA is to the square on EF--that is, as the circle on DA is to the circle on EF. And it is always so wherever the point E. Therefore, since proportional magnitudes of two types are suspended from the same points, as was required in the preceding lemma, all the magnitudes of the first type together (that is, all the lines of the trilineum ABCD, or the trilineum itself) will have the same point of equilibrium as all the magnitudes of the second type together (that is, all the circles of the cone ACD, or the cone itself) have. Moreover, the cone balances at equilibrium from the point which divides CD in such a way that the part toward C is three times the remaining part, since the line DA is perpendicular to the horizontal.*^{4} Therefore, the trilineum ABCD also balances at equilibrium from the same point. *Quod erat propositum, &c.*

#### Proposition 14.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter DE is understood to be perpendicular to the horizontal. Let CF and AD be tangents, and let AF in fact be parallel to the diameter.

Let FH then be assumed as 1/4 of the whole FC. The mixed trilineum ABCF will balance at equilibrium from the point H (by the preceding Lemma). Also, let FI be taken as 1/3 of the whole FC. The whole triangle AFC will balance at equilibrium from I.^{5} Clearly the parabola will balance at equilibrium from D, since it has a center on the diameter. Therefore, the trilineum ABCF will be reciprocally to the parabola itself as DI is to IH--namely, 2 to 1. (In particular, if FC is 12 of some parts, FD is 6, FI is clearly 4, and FH is 3. Therefore, DI is 2 and IH is 1.) Consequently, by componendo, the whole triangle AFC will be three times the parabola. The remainder of the quadrature is completed as was done in Proposition 9. *Quod erat &c.*

#### In another way.

*With the same things assumed as above, let FH be supposed 1/4 part of the whole FC. The mixed trilineum ABCF will balance at equilibrium at the point H. Also, let FI be supposed 1/3 part of FD itself. Then indeed the triangle FDA will balance at equilibrium from the point I. Clearly, the mixed trilineum ABCD will balance at equilibrium from the point D. (For the entire triangle ADC balances at equilibrium from the point D, and the removed parabola balances at equilibrium from the same point D. Therefore, it is also necessary that the remaining trilineum ABCD balance at equilibrium from the point D.) And so the triangle FDA will be to the trilineum ABCD reciprocally as DH is to HI--namely, as 3 is to 1. By conversion of the ratio, the triangle ADC will be to the parabola as 3 is to 2, or as 6 is to 4. Wherefore, the parabola will be to the triangle ABC as 4 is to 3--namely, 4/3. **Quod erat propositum demonstrare &c.*

*With the following observation on geometric progressions said first, we may approach the quadrature of the parabola from yet another supposition. *

#### Lemma 24.

If two straight lines should intersect each other and between these should be described a certain constant flexilineum from alternately parallel lines, all lines which are parallel among themselves will be in continuous proportion.

*Let two straight lines AB and CB intersect each other at point B, and between these let a flexilineum CADEFG etc. be described in such a way that CA, DE, FG, etc. are parallel among themselves, and likewise AD, EF, and the remaining in turn are assumed parallel among themselves. I say that AC, ED, and GF are in continuous proportion. *

*Indeed, since they are parallel, as AC is to ED, so is AB to BE, or DB to BF--that is, ED to GF. (Euclid VI.2 and VI.4.) **Constat ergo quod propositum fuerat*.

#### Lemma 25.

With two straight lines intersecting each other as shown above, if the two lines AC and DE are parallel between themselves and with CD joined, let it be conceived that the flexilineum ACDE is continued *in infinitum* all the way to the point B of intersection. I say that in a flexilineum of this kind are to a hair each and all the terms which are in the progression of the proportion of AC to DE continued *in infinitum*.^{6}

*Let F be assumed equal to AC itself and G equal to DE. Also, let the proportion of F to G be conceived as continued in its own infinitely many terms from F to H. *

*Now if it is possible, let there be some term or terms in the progression from F to H which are not found in the flexilineum and let I be greatest term of those terms which, though they are in the progression FH, are not in the flexilineum. Therefore, the term L preceding such an I will be in the flexilineum. Let it be MN. Since L is to I as F is to G, or as AC is to DE, or as NM is to PO, following next, and L and NM are equal, I and PO will also be equal. Therefore, the term I, which was assumed to not be in the flexilineum, was found in the same. *

*We would demonstrate in completely the same method that there is no term in the flexilineum which is not also in the progression FH etc. We would then conclude that all the terms in the flexilineum are precisely the terms of the proportion of AC to DE continued **in infinitum*, since it was demonstrated that no term which is in the progression FH is missed in the flexilineum nor is any term which would not also be found in the progression FH present in the flexilineum.

#### Lemma 26.

With infinitely many straight lines of a greater than inequality assumed in continuous proportion, to find a straight line which is equal to all the aforementioned lines together.

*Let A and B be the first two lines of the given progression, in which CD is supposed equal to the greater A and EF to the lesser B. Let CD and EF be parallel, and let DF and CE, which intersect by necessity, be joined. Thus let them intersect at point G and, with CF drawn, let GL be parallel to the CF. *

*I say that the line DL is equal to all the infinite terms of the progression ABM taken together. *

*Indeed, let it be conceived that the flexilineum DCFE etc. is continued **in infinitum* all the way to the point G. All the lines will be in the flexilineum itself, or the terms of the given progression ABM.

*Now let HE, NI, and the remaining lines parallel to them be extended all the way to DL. EF will be equal to CP itself, HI equal to PQ, and NO equal to QR, and so on one at a time (by Euclid I.34). Indeed, each line which is in the flexilineum will have its own corresponding small part on the line DL equal to itself until the flexilineum should arrive at the ultimate point G. At that time, moreover, there will not be anything from the flexilineum or from the line DL that will be left over, but the flexilineum itself as much as the line DL will also have been completely used up. Indeed, GL itself, which is drawn from the final point G of the flexilineum, is the last of all the parallels which are drawn all the way to DL. Therefore, all the lines of the flexilineum together, of which the first is CD, taken alternately (that is, all the lines of the progression ABM) are equal to all the small parts of the line DL taken together (that is, DL itself). **Quod erat ostendendum &c.*

#### Lemma 27.

With infinite magnitudes supposed in a continuous geometric proportion of a greater than inequality, the first magnitude will be the mean proportional between the first difference and the aggregate of all the magnitudes.

*Indeed, with the preceding construction assumed, let FV be drawn parallel to the GC. Then DV will be the first difference. But DV is to the first magnitude DC as FD is to DG (Euclid VI.4)--that is, as DC is to the aggregate DL of all of them. **Quod erat demonstrandum &c.*

#### Comment.

*We will not hesitate to assert that this is also true for numbers and any sort of magnitudes. We will convey a demonstration even more universal, especially since it is quite brief. The conclusion of this truth, when it had been contrived in passing by us for the most celebrated Cavalieri, also itself established the same Theorem with the following demonstration, which has been already supplied by us in the first proof. *

*This is said in advance. But if there will be however many magnitudes, either finite or infinite in number, the antecedents of which are always larger than the consequents, the first magnitude of all will be equal to all the differences taken together with the smallest magnitude itself. *

*This was known among geometers and is demonstrated as was done by us in Lemma 15, where we show that the parallelogram AE is equal to all the differences between the following parallelograms and the smallest parallelogram OC. *

*Now let magnitudes be supposed infinite in number in a continuous geometric proportion of a greater than inequality. It is evident that the smallest magnitude of all either will or will not be a point. Therefore, in this case the first magnitude will be equal to all such differences. *

*Moreover, since the magnitudes are assumed in a continuous geometric proportion, the differences will also be proportional in the same ratio and therefore (with a conversion having been done) as the first difference is to the first magnitude, so will the second difference be to the second magnitude, and thus always. Consequently, as one is to one, so summarily will all be to all--namely as the first difference is to the first magnitude, so are all the differences together (that is, the first magnitude itself) to all the magnitudes together. Therefore, it stands that the first magnitude is the mean proportion between the first difference and the aggregate of all of them. *

#### Proposition 15.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola in which is inscribed the triangle ABC. I say that the parabola is 4/3 of the triangle ABC.

Indeed, let the two triangles ADB and BEC also be inscribed in the remaining portions ADB and BEC of the parabola. The triangle ABC will be four times the two triangles ADB and BEC together (Lemma 7). Let four inscribed triangles also be conceived in the remaining small parts AD, DB, BE, and EC. Both triangles ADB and BEC together will be four times the aforementioned subsequent triangles together (Lemma 7), and always in this manner. Therefore, the parabola is nothing other than a certain aggregate of magnitudes infinite in number in a 4 to 1 proportion, of which the first is the triangle ABC, and indeed the second consists of the two triangles ADB and BEC. Therefore, the first magnitude ABC will be the mean proportional between the first difference and the aggregate of all of them--namely, the parabola.

Consequently, suppose the triangle ABC is 4. Therefore, the two triangles ADB and BEC together are 1, and the first difference (of course between 4 and 1) is 3. Therefore, the aggregate of all the infinite magnitudes (namely the parabola itself) will be (by lemma 27) to the first magnitude (that is, to the inscribed triangle ABC) as the first magnitude itself is to the first difference--clearly, as 4 is to 3. Namely, 4/3. *Quod erat propositum demonstrare &c.*

#### In Another Way.

*Let ABC be a parabola whose diameter is DB, let AD and CD be tangents to the base, and also let EF be tangent to the. Moreover, in the remaining trilinei ABE and BCF let the two triangles GEH and IFL be inscribed (as was required for the construction in the third and fourth lemmas). Likewise, let four triangles be conceived in the four remaining mixed trilinei, and always in this way. So the entire trilineum ABCD will be nothing other than a certain aggregate of magnitudes infinite in multitude in a 4 to 1 proportion (Corollary 1 of Lemma 3), of which the first is the triangle EDF, the second in truth consists of the two triangles GEH and IFL, and the third in truth from the following four, etc. Therefore, the aggregate of all (namely, the mixed trilineum ABCD) will be to the first magnitude (namely, to the triangle EDF) as the first magnitude itself is to the first difference (Lemma 27)--clearly, as 4 is to 3. *

*Thus, since the trilineum ABCD is to the triangle EDF as 4 is to 3, the same trilineum will be to the triangle ADC as 4 is to 12. Therefore, the parabola will be to the triangle ADC as 8 is to 12 and to its own inscribed triangle as 8 to 6--namely, 4/3. **Quod erat demonstrandum &c.*

#### Lemma 28.

If straight lines AB, CD, EF, etc. infinite in number are in a continuous geometric proportion of a greater than inequality and moreover another progression BG, DH, FI, etc. is supposed so that just as the first AB is to the first BG, so the second CD is to the second DH, and so the third EF is to the third FI, and thus always, I say that the entire aggregate of the progression AB, CD, EF, etc. is to the aggregate of the progression BG,DH, FI as AB is to BG.

*Let all the terms of the two progressions be conceived to be in the flexilinei (in like manner with Lemma 25). With AD and GD joined, let OL be drawn parallel to AD itself and OM parallel to DG itself. BL will be equal to all the infinite terms AB, CD, EF, etc. (Lemma 26), and indeed OM will be equal to all the infinite terms of the remaining progression BG, DH, and FI (Lemma 26). *

*Now as LB is to BA, so is OB to BD--that is, MB to BG (Euclid VI.4). Therefore, by permutando, the aggregate LB is to the aggregate BM as AB is to BG--namely, as one magnitude is to the other. **Quod erat &c.*

*This theorem had been able to be substituted for the demonstration in Proposition 12 of Book 5 of Euclid. Indeed, it is one and the same with the Theorem in the aforementioned Proposition. But in fact, since almost all are of the opinion that Euclid there supposed a multitude finite in number, we have determined with the help of flexilinei that, *

#### Proposition 16.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter is DE, whose tangents at the base are AD and CD, and whose tangent through the vertex is in fact FBG. Let ABC be the inscribed triangle. I say that the parabola is 4/3 of the triangle ABC.

For since the EB is equal to the BD (because it is a parabola),the line AC is in fact twice the line FG and the inscribed triangle ABC will be twice the triangle FDG contained by the tangents. And this is also always true around the remaining parabolic parts AIB and BOC. (Indeed, AIB is a parabola whose tangents at the base are AF and BF. Therefore, the inscribed triangle AIB will be twice the tangent triangle LFM. The same is also true for the other part. Therefore, the two triangles AIB and BOC together are twice the two triangles LFM and NGP together.) Therefore, since the two progressions are both in a continuous proportion of magnitudes infinite in multitude (namely, the one inside the parabola whose first term is the triangle ABC and second term is the two triangles AIB and BOC, together, etc, and also the other progression outside the parabola whose first term is namely the triangle FDG and second moreover is the triangles LFM and NGP together, etc.) and since the individual terms of the progression inside the parabola are twice the individual terms of the progression outside the parabola, therefore the entire aggregate of the first progression will be twice the entire aggregate of the second progression (Lemma 28). In particular, the parabola itself will be twice the mixed trilineum ABCD. Therefore, by componendo and conversion of the ratio, the triangle ADC will be 3/2 of its own parabola--namely, as 6 is to 4. Therefore, the parabola will be to the triangle ABC as 4 is to 3--clearly, 4/3. *Quod erat ostendendum &c.*

*With the aid of infinitesimals the quadrature of the parabola is able to be gained with yet other results assumed. We suppose Propositions 14 and 25 which Archimedes demonstrated in the book **on Spiral lines* with a lemma of this kind set out beforehand.

#### Lemma 29.

If a first magnitude is to a second as a third is to a fourth, and thus however often it will have been pleasing, and if all the firsts and also all the thirds are proportional in the same order, then all the firsts together will be to all the seconds together as all the thirds together are to all the fourths together.

*Let the first A be to the second B as the third C is to the fourth D, and E to F as G to H, and thus however often it will have been pleasing. And let all the firsts A, E, I, etc. and all the thirds C, G, M etc. be proportional in order--namely, as A is to E so should C be to G. Moreover, as A is to I, so should C be to M, etc., and thus always. I say that all the firsts A, E, I etc. together are to all the seconds B, F, L etc. together as all the thirds C, G, M etc. together are to all the fourths D, H, N etc. together. *

*Let O, P, Q etc. be taken as individuals equal to the first of the firsts--that is, to A itself--and let there be just as many as there are all the firsts A, E, I, etc. Likewise, let R, S, T etc. be assumed to be just as many as there are all the thirds, and let the individuals R, S, T be equal to the first of the thirds--namely, C itself. *

*Now on account of equality, as O is to A so will R be to C. Further, since P is equal to A itself, and S to C itself (because of the supposition) as P is to E so will S be to G. And this always. All O, P, Q are equal, and likewise all R, S, T are equal. Therefore, all O, P, Q, etc. together are to all A, E, I, etc. as all R, S, T together are to all C, G, M together (Lemma 18). Finally, by convertendo, all R, S, T are to all O, P, Q as all C, G, M are to all R, S, T. Remember this. *

*Since indeed as O is to A so is R to C and as A is to B so is C to D, O will be to B as R is to D (by equality). For completely the same reason, we conclude that as P is to F so is S to H (by equality), and thus concerning the rest. Therefore, all O, P, Q etc. together will be to all B, F, L etc. as all R, S, T etc. together are to all D, H, N etc. (Lemma 18). Wherefore, by equality, all A, E, I etc. will be to all B, F, L etc. as all C, G, M etc. are to all D, H, N etc. **Quod erat ostendendum*.

#### Proposition 17.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose tangent is AE. Indeed, let the diameter be parallel to CE, and let some FD be drawn parallel to CE itself. EC will be to FB in length as EA is to AF^{2} (because it is a parabola), or as EC^{2} is to FD. Therefore EC, FD, and FB will be in a continuous proportion.

Then let the intervals AC and AD become two circles with center A. Let the beginning of a spiral be placed on the semidiameter AC, and let the spiral itself be AGC.

So DF will be to FB as CE is to DF, or as CA is to AD--that is, as CA is to AG, or as the entire circumference CLHC is to the arc CLH (Proposition 14 of *On Spiral Lines*). That is, as the entire circumference DPGD is to the arc DPG. Indeed, it will always be thus wherever the point D is taken. All the first magnitudes and likewise all the thirds, are proportional in the way in which they ought to be (as we will show below). Moreover, all the firsts together (namely, the triangle AEC) will be to all the seconds together (namely, the mixed trilineum ABCD) as all the thirds together (namely, the circle CLH) are to all the fourths together (that is, to the remainder of the circle itself with the region inside the spiral CAGC subtracted) by the preceding lemma. Moreover, the circle CLH is 3/2 of the aforementioned space with the region inside the spiral subtracted (Proposition 25 of *On Spiral Lines*). Therefore, the triangle ACE will also be 3/2 of the mixed trilineum ABCE. By conversion of the ratio, the triangle ACE will be three times the parabola ABC. The remainder of the quadrature can be completed as in Proposition 9. *Propositione factum est*.

*Moreover, we will now show what was assumed--namely, that all the firsts and all the third magnitudes are proportional in the way that is required in the preceding lemma. *

*Let some MO be drawn parallel to FD itself as in the previous figure, and let us suppose that the FD is the first of the firsts and that the circumference DPG is itself first of the thirds. Therefore, DF will be to OM as DA is to AO, or as the circumference DPG is to the circumference whose semidiameter is AO etc. And thus always. **Quod oportebat &c.*

*We will also square the parabola by a way as of yet untried--of course, with its center of gravity found with the prior help of indivisibles. Moreover, we suppose a lemma which Archimedes showed in the Book 2 of ** on the Equilibrium of Planes*--that is, that the centers of gravity of parabolas divide their own own diameters in the same proportion.

#### Lemma 30.

The center of gravity of a parabola divides the diameter so that the part ending at the vertex is 3/2 of the remainder.^{7}

*Let ABC be a cone whose base is AMC, axis is BC, and the axial triangle is in fact ABC. Let the cone be cut by the plane EFG as was demanded in Proposition 11 of ** The Conics, Book I*. The section will be that which is called a parabola, and its diameter will be FH. Now let the center of gravity of the parabola EFG be some point, call it I. It must be shown that the line FI is 3/2 of the IH.

*Let the line AIL be drawn through the point I, and let the cone be cut by another plane MNO parallel to the EFG. The section on MNO will be a parabola, and its center of gravity will be P. (For since they are parabolas, as FI is to IH so is NP to PR. But I is assumed the center of gravity of the parabola EFG. Therefore, by Proposition 7 of of **On the Equilibrium of Planes, Book II*, P will be the center of gravity of the parabola MNO.) And thus always, wherever the plane MNO. Therefore, the centers of gravity of all the parabolas which are in the cone ABC are found individually on the line AL. Wherefore, the common center of gravity of all the same aforementioned parabolas together will also be on the line AL. Moreover, all the parabolas are the same as the cone itself. Therefore the center of the cone is on the line AL, but since it is also on the axis BD, the center of the cone will be on the common point of intersection S. Therefore, BS will be three times SD itself.

*Let the line DQ be drawn from the center of the base parallel to AL. CQ and QL will be equal (Euclid VI.2). Moreover, since BS is three times SD (because S is the center of the cone), BL will also be three times LQ (Euclid VI.2). Therefore, BL is 3/2 of LC. Wherefore, FI will also be 3/2 of IH. **Quod erat propositum &c.*

#### Proposition 18.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter is BD, and indeed let the inscribed triangle be ABC. I say that the parabola is 4/3 of the triangle ABC.

Let AD and DC be divided in half at the points E and F, and let EG and FH be drawn parallel to the diameter. These will be the diameters of the segments AGB and BHC. Let the centers of gravity of the aforementioned segments be O and N. GO and HN will each be 3/2 of the remainders OI and NL (by the preceding lemma). Let ON be joined. The common center of gravity of the two segments will be on ON itself (by Proposition 8 of *On the Equilibrium of Planes, Book I*). But it is also on BD. (For as the center of the whole parabola is on BD, so also is the center of the triangle ABC.) Wherefore, the point P will be the center of the segments AGB and BHC. Assume BD is 60 parts. GE will be 45 parts (since it is 3/4 of BD), the IE will be 30 parts, and the EO--that is, DP--36 parts. Let Q be the center of gravity of the triangle ABC; DQ will be 20 parts. Let R be the center of the parabola; RD itself will be 24 parts (by the preceding lemma). Therefore, PR will be 12 parts and RQ 4 parts. But as PR is to RQ so reciprocally is the triangle ABC to the two segments AGB and BHC. Wherefore, the triangle ABC will be to the two portions AGB and BHC as 12 is to 4--namely, as 3 is to 1. By componendo and conversion of the ratio, the parabola ABC will be to the triangle inscribed in it as 4 is to 3--namely, 4/3. *Quod erat propositum &c.*

*We will undertake the quadrature of the parabola by a hitherto new calculation, with the following lemma assumed, which indeed was said to have come forth from the Cavalerian School. Indeed, it was devoted to the measurement of a certain solid arising from the parabola itself having been revolved around its axis. Moreover, the lemma is of a type from the author Ionas Antonio Roccha, a distinguished geometer.*^{8}

#### Lemma 31.

If a plane figure is balanced above some straight line of its own which divides the figure itself, the moments of the segments of the figure are as the solids of revolution described by the segments having been revolved around the dividing line.

*Let ACDBFE be a plane figure which the straight line AB divides, and let the figure be imagined to be balanced above the line AB. I say that the moment of the segment ACDB is to the moment of the segment AEFB as the solid of revolution arising from the revolution of the segment ACDB around the axis AB is to the solid of revolution arising from the revolution of the remaining segment around the same axis of revolution. *

*For with two given points H and I assumed on the line AB, let the lines CE and DF be drawn through H and I perpendicular to the AB, and let the segments DH and HF be cut in half at the points L and M. *

*Therefore the moment of the line DH will have a ratio to the moment of the line HF compounded from the ratio of the magnitudes DH to HF, and from the ratio of the distances LH to HM--or DH to HF.*^{9} Therefore, the moment of the line DH will be to the moment of the line HF as the square on DH is to the square on HF.

*It will be shown in the same manner that the moment of the line CI is to the moment of the line IE as the square on CI is to the square on IE, and thus always. *

*Further, the moment on DH is to the moment on CI (on account of the same reason as above) as the square on DH is to the square on CI, and in this way always. Therefore, (by Lemma 29) all the first magnitudes together, (namely, all the moments of the figure ACDB) will be to all the seconds together (namely, to all the moments of the remaining figure AEFB) as all the thirds together (namely, all the squares of the figure ACDB) are to all the squares of the remaining figure, or as all the circles of the figure ACDB (namely, the circular solid described by its rotation around the axis AB) are to all the circles of the remaining figure AEFB (namely, to the circular solid arising from its revolution around the same axis AB). **Quod erat ostendendum &c.*

*With this having been set forth (which, as we indeed declared thoroughly above, was undertaken by others, was introduced here as another's, and which I believe was not well known until now), we will square the parabola, with the demonstration, which is proved in many ways, assuming that the Cylinder is twice its own inscribed parabolic conoid.*^{10}

#### Proposition 19.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABCD be a semiparabola, around which is a rectangle DE. Let a point F be assumed such that AF is to FD as 5 is to 3, and let FG be drawn parallel to the diameter. The center of gravity of the semiparabola will be on the FG (by Lemma 11). Let it be some point, call it I, let LIM be drawn parallel to AD through I, and let IN be taken equal to IM itself. Also, let PQ (wherever it may fall) be understood as extended parallel to the diameter CD in such a way that the rectangular parallelogram DP is equal to the semiparabola itself. Then let a rectangle DR be conceived applied to the line CD in such a way that it balances at equilibrium with the semiparabola with the balance above the line CD. Let the center of the aforementioned rectangle be the point S and, with TSX drawn parallel to the AD, let the line IS be joined.

Now it is clear from the lemma above that the cylinder made from the rectangle DR revolved around the axis DC will be equal to the parabolic conoid made from the rotation of the semiparabola ACD around the same axis of revolution CD, since the moments of the plane figures are assumed equal. Therefore, the cylinder made from the rectangle DR will be 1/2 of the cylinder made from the rectangle DE, and consequently the square on TX will be 1/2 of the square on ML. (For cylinders of equal height are among themselves as the squares of the bases.) Remember this.

In fact, MN is to TX as IM is to TS (Euclid VI.4)--for they are halves of the same things), or as IV is to VS--in particular, (because the plane figures balance at equilibrium above the line CD, or from the point V) reciprocally as the rectangle DR is to the semiparabola (or to the rectangle DP equal to the semiparabola itself), or as their bases TX to MO. Therefore, TX is the mean proportional between MN and MO. Wherefore, the rectangle NMO will be 1/2 of the square on LM, since it is equal to the square on TX.

Clearly the ratio of the square on LM to the rectangle NMO is composed from the ratio of LM to MN (which is 4/3 by construction, for we assumed the point F is such that AF is to FD as 5 is to 3) and from the ratio of LM to MO, which indeed was unknown, but now shows itself as 3/2 by necessity. For the ratio of 2 to 1 is composed from 4/3 and 3/2, as also was made known by the Cantorians themselves, as was seen in the three numbers 4,3,2.

Therefore, the rectangle DE to DP itself, or to the semiparabola, will be 3/2, and the semiparabola to the triangle ACD will be 4/3. *Quod erat ostendendum &c.*

#### Lemma 32.

Let ABC be a parabola with base AC and tangent CD. Let AD be parallel to the diameter. With some point E assumed, let EF be drawn parallel to the diameter. I say that as FE is to EB, so is CA to AE.

For DA is lengthwise to the FB as DC^{2} is to CF^{2}, or as DA^{2} is to FE^{2} (because it is a parabola). Therefore, DA, FE, and FB are in a continuous ratio. Remember this.

Now as AC is to CE so is AD to EF, or EF to FB. By conversion of the ratio, as CA is to AE so is FE to EB. *Quod erat ostendend. &c.*

#### Lemma 33.

Any parabola is equal to two parabolas taken together which indeed have a base equal to that one, a diameter in fact half, and which are equally inclined.

*Let ABC be a parabola whose diameter is BH. Let the two other parabolas AEC and AGC be on the same base and indeed let the diameters HE and HG both be 1/2 of the diameter HB, but inclined equally to the base. I say that the parabola ABC is equal to the figure AECG. *

*For let some point be taken on the base AC, and let it be M. With PMN drawn parallel to the diameter BH, BH will be to NM as the rectangle AHC is to the rectangle AMC, or as the line HE is to the line MO. By permutando, as BH is to HE so will NM be to MO. Wherefore, NM will be twice the MO. It will be shown in completely the same manner that NM is also twice the MP. Therefore, the entire NM is equal to OP itself. And this always. On that account, all lines of the figure ABC together (namely, the parabola ABC itself) will be equal to all the lines of the figure AECG together (namely, to the two parabolas AEC and AGC). **Quod erat &c.*

#### Proposition 20.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a parabola whose diameter BE is conceived as perpendicular to the horizontal, and let the parabola itself be inverted. Let CA be extended to D, so that CA and AD are equal, and let DC be a balance whose fulcrum is A. Let CF be drawn tangent to the parabola and AF parallel to the diameter EB. Suppose GH is equal to AC itself, and let GH be divided in half at I. Let IL and IM both be half of the line EB and be inclined equally to the base as EB itself is to AC. Also, let the two parabolas GLH and GMH be constructed, which (by the preceding lemma) together will be equal to the parabola ABC, and let the figure GLHM be suspended from the point D.

Now let the points O and N be taken equally in distance from the points I and E, respectively. With NQ drawn parallel to EB and ROS to LM, NP will be equal to RS itself, as in the preceding lemma.

Now QN is to RS as QN is to NP (because of equality), or reciprocally as DA is to AN (by Lemma 32). Therefore, the lines QN and RS balance at equilibrium, and thus always. Therefore, all the lines of the triangle AFC together (namely, the triangle itself) balance at equilibrium all the lines of the figure GLHM together (namely, the figure GLHM itself).

Let AV be 1/3 of the entire AC. It is clear that if the line from V is sent down parallel to AF itself, the center of gravity of the triangle AFC will be on it, and the line will be perpendicular to the horizontal. Therefore, the triangle AFC will be suspended centrally from the point V, and the triangle AFC will reciprocally be to the region GLHM as DA is to AV--namely, 3 to 1.

Moreover, since the region GLHM is equal to the parabola ABC, the triangle AFC will also be three times the parabola ABC.

*The remainder of the quadrature is solved as was done in Proposition 9. **Quod &c.*

#### Proposition 21.

The parabola is 4/3 of the triangle that has the same base and height.

Let ABC be a semiparabola whose diameter is CE and ordinate is AE, and let CD be tangent. Let the parallelogram AECD be completed. It is clear that all the lines of the mixed trilineum DABC which are in fact parallel to the diameter are in the same ratio among themselves as all of the circles of a cone which has axis DC and vertex C. Therefore, the center of gravity of all the lines of the trilineum DABC, will be on that line which divides the balance DC just as the center of gravity of the cone divides the same balance--namely, so that the part nearest to C is 3 times the remainder (by Lemma 22). Therefore, let CF be made 3 times FD itself, and let FM be drawn parallel to CE. The center of gravity of the trilineum DABC, wherever it may be, will be on the line FM.

Likewise, all lines which are drawn parallel to the diameter of the semiparabola ABCE are in the same ratio among themselves which are all circles of some hemisphere whose axis is AE and vertex clearly is A (by Lemma 22). Therefore, the center of gravity of all the lines suspended from the balance AE, or from the semiparabola itself, will be on that line which thus divides the balance AE as the center of gravity of the hemisphere divides the balance. Namely, so that the part terminated at A is to the remainder as 5 is to 3.^{11} Therefore, let AI be to IE as 5 is to 3, and let IH be drawn parallel to CE. The center of the parabola, wherever it is, will be on the line IH. Finally, let GL be drawn which cuts in half the sides AE and DC. The center of gravity of the parallelogram DE, which is O, will be on GL. Suppose that the center of gravity of the semiparabola is some point P, and let PO be extended onto N. N will be the center of gravity of the trilineum DABC. Now the semiparabola is to the trilineum as NO is to OP, or as ML is to LI--namely, as 2 is to 1. (For by construction the whole AE is 8 of some kind of parts, AM is 2 of such parts, ML is 2, LI is 1, and the remaining IE will be 3.) Therefore, the semiparabola will be to the parallelogram as 2 is to 3, or as 4 is to 6, and the semiparabola is to its inscribed triangle as 4 is to 3--namely, 4/3. *Quod &c.*