On the transmutation of equal rectilinear surfaces into various curvilinear figures, and likewise the converse.

b a; L — p o m n; c

a b is squarable in itself when its four segments are filled with the four larger half-segments n m o p; and then the large segment placed above the void is given back, whence m o was removed, and it will remain like the small quadrature above.

b a - o m n

a b is squarable by filling the empty segments with the dotted segments above and at the side.

nocb-vSrt -ma

It is squared by putting the lunes b n into r S, and there will remain the triangle a m r S c o, made whole again in its voids r S.

-a-cb

a is worth b c. It is squared by filling the concavity of the two lunes with their convexity.

e a d - c b

a is worth b c with its supplements d e. It is squared by filling the oval voids with the two lunes d e and with the dotted segments.

+ a-cmnb

It is squared by filling the voids n m with the four convexities of a b c.

+ c b - a

It is squared by filling the four concavities of a with the 4 lunes b c, converted into 4 segments of like value.

cab

a is worth b c. It is squared by filling half of each of the two voids with half of each of the two segments.

o n — q p

It is squared by putting the segments n o into p q.

n-b-o-a-qmp

n m o p q is worth the greatest triangle of the semicircle, from which the two greatest segments are removed.

b a

It is squared by removing the 2 upper segments of the circle and giving back the value of the two empty lower lunes.

And it is also squared by placing a b above, embracing the convexity of the two segments with their concavity; and it will stand as above. — False.

+ —b a -d c

a b c d are equal to one another and are squarable in themselves by taking and giving back part for part.

a-m n

It is squared by putting the two lunes inside the segments that are enclosed within the greatest triangle.

It is squared by filling the void of the four segments with the two dotted segments, and the rectilinear triangle will be made.

b a

It is squared by filling the three voids with the two dotted segments a b.

It is squared by filling the oval concavity with the four lunes from outside, and the greatest triangle is made whole again.

It is squared with the four lunes, filling with them the four segments that are nearest the underside of the square.

ba-n; c-emd

If equal parts are taken from equal things, the remainder will be equal. Take therefore a b, segments of the semicircle a n b, and take d c e from d c e m, and there will remain n equal to m.

It is squared by filling the four empty lunes inside the greatest triangle with the four lunes that lie outside that greatest triangle.

It is squared by removing the two dotted segments made on the convexity of the lune, and filling the greatest empty segment with their value.

b a - m n

a is worth two segments, or if you will two lunes. Fill n m with a, and with b fill the two empty segments, and let the lunes fill the remainder.

It is squared by filling the voids of the greatest triangle with the four lunes from outside.

It is squared by putting back inside the greatest triangle what is included between the circumferential line and it.

It is squared by putting back inside the quadrangle the four lunes that are outside it.

It is squared by filling the eight segments and removing their value by removing the two greatest segments.

b a - no

The lune-shaped segments a b fill the six smaller segments, and the two o n remain fixed to the middle quadrangle.

b a

a b is worth the black field, that is, the eight empty segments. It is squared by filling the black field with the two segments a b.

b — c a

It is squared by filling the greatest square with the segments a b c.

The lune is squared by putting the convex part into the concave, and the like is done with what remains outside the lune. But it is false, because the lune is sided (angular).

It is squared by filling the voids of the greatest triangle with what is outside it.

It is squared by filling the two concavities above with the two convexities below.

It is squared by filling the six empty middle segments with the 4 lunes and two segments above.

It is squared by filling the inner lunes with the ovals, that is, the halves of the width of the four outer ovals.

c -b a

a b is worth c. It is squared by putting the convex parts into the concave ones.

m n

The four black half-segments are worth the two greatest outer segments. It is squared by filling the triangle with n m.

r c

It is squared by filling the greatest triangle with c r.

It is squared by putting the convexity back into the concavity.

It is squared by putting the convexity back into the dotted concavity.

It is squared by filling the two lunes and the oval figure, which in all are worth 4 smaller segments, and by removing from that combination the greatest segment above.

It is squared by putting inside the quadrilateral what is outside it, that is, the four lunes: they are worth the 4 segments.

It is squared by filling the next-to-last segments above with the two lunes; and then you remove the two greatest half-segments at the side, and with those the 4 segments that remained are filled.

It is squared by putting the convex part back into the concave part.

It is squared by putting back inside the greatest triangle what is outside.

It is squared by filling the concavity above with the two dotted lateral segments.

It is squared by filling the two concavities of the two squares with the two lunes above.

It is squared by removing the 4 segments of the two semicircles, and with those you fill the greatest segment below.

It is squared when you fill the lower segments with the upper lunes.

It is squared by putting back inside the greatest triangle what lies outside it.

It is squared by putting inside the greatest triangle the two lunes that lie outside that triangle.

It is squared by putting the lunes that are outside the triangle beneath its base.

It is squared by putting inside the greatest triangle the two convex segments that project outside that triangle.

It is squared by filling the inner concavity with the two outer convexities.

It is squared by removing the 4 outer convexities and putting them back into the two inner concavities.

It is squared by filling the two empty segments with the two lunes that lie outside the greatest triangle.

It is squared by filling the two empty segments that lie inside the greatest triangle with the two lunes that are outside that triangle.

It is squared by filling inside the greatest triangle with what is outside that greatest triangle.

It is squared by filling the greatest triangle with what is outside that triangle.

b a -h

It is squared by filling, with the two lunes a b, the 4 inner half-segments included in the contact of the curvilinear triangle h.

It is squared by filling the triangle with the two outer lunes.

It is squared by filling the triangle with the four outer lunes.

It is squared by filling the left middle one with the greatest segment, and with the lune outside the greatest triangle the middle one is filled.

It is squared by filling the oval inside the smaller semicircle with the dotted segment.

It is squared by filling between the two lunes with the 2 lunes and 2 segments that are at the side, and beneath the 2 lunes it is filled with their convexity.

It is squared by putting the 4 outer lunes inside the greatest triangle in all the voids, small and large.

It is squared by putting the two dotted segments that project outside the lune into the 4 smaller segments, and the two lunes are put outside the lune.

It is squared by filling the lune with the two lunes and two segments that lie outside it, and the convexity of the lune with the two segments below.

It is squared by filling the lune with the two lunes, and the two points of the lune fill its two lower inner segments.

It is squared by filling the upper semicircle with its convex part, and the two small lower segments are filled with the two horns of the moon.

It is squared by filling the four smaller lower segments with the two convexities of the lunar horns, and the upper segments are filled by the two lunes.

It is squared by filling the 2 lower segments with the 2 lateral lunes, and the 2 upper lunar horns fill the 2 small segments that are beneath the square.

It is squared by filling beneath the lune with the two lateral segments, and inside the lune it is filled with the two convexities of the lune.

It is squared by filling beneath the lune with the two lateral segments, and beneath the square it is filled with the two dotted horns of the lune that exceed the greatest triangle.

It is squared by putting the 4 lunes that are outside the greatest triangle into the 4 empty segments inside that triangle.

It is squared by filling the 4 smaller segments that are inside the greatest triangle with the 2 lunes that lie outside that triangle.

It is squared by filling the two lower lunes with the supplements. Let the like be done to fill the 8 small segments.

It is squared by filling the empty lunes that are beneath the circle with the full lunes that lie outside that circle; and the 4 small segments that lie in the circle are filled with the upper convexity of that circle.

It is squared by filling 6 of the 8 empty segments with the 4 lunes and two segments that lie on the convex part of the two dotted lunar horns.

The lunes are squarable in themselves, because they are those of the double circles of Zenophon, and the remainder is squarable.

It is squared by filling the lune-shaped voids with the lune-shaped surfaces, and by filling the 2 small voids of the small inner segments with the convex lunar horns.

It is squared by filling the lune-shaped spaces that are inside the greatest triangle with the full lunes that lie outside that greatest triangle, and the 4 small segments are filled with the convex parts of the semicircle.

It is squared by filling the concave parts with the convex parts of the parallel semicircles.

The segments put into the empty segments square the empty circle.

The quadrature of this circle is done by putting the half oval figures that are outside the square into the voids of that square.

Regular figures made inside the semicircle. All these semicircles are equal to one another, and equal are the principal parts into which they are divided; and again equal are the parts that each of the two aforesaid parts has. Therefore the remainders of each part are equal, and so on. Every curvilinear surface to which a square equal to it is given is also squarable in itself by its parts.

If you take equal parts from equal things, the remainders are equal to one another. If two surfaces are in double proportion and are entirely superimposed one on the other, the excess of the larger is worth the smaller.

a-So-nbd

a is worth b. It is squared by filling the voids o S with the dotted segment a.

c n o b - a

a is worth b c. It is squared by filling the voids o n with the two dotted segments b c.

b — m — n a o

a is worth b. It is squared by filling the voids o m n with the whole segment b.

S-m-cb-tao

a is worth b c. It is squared by filling the voids o n m with the segment S.

d - b a - o n - c

a b c is worth d. It is squared by filling the voids n o with the segment d.

e b a d - p o - c

a b c is worth d e. It is squared by filling o p with the d e half-segments.

c - m a - b n

a b is worth c. It is squared by filling the spaces n m with the dotted segment c.

c bo— a— rS— a

a a is worth b o c with its supplements; o goes into S and into r, and it remains squared.

d - c r o b - a

a is worth b c d. It is squared by filling the voids o n with the dotted half-segments b c.

c a b — p o

a is worth b c. It is squared by filling o p with the two dotted half-segments b c.

c a b - o n

a is worth b c. It is squared by filling the void n o with the two dotted half-segments n o.

a - m n - d c b

a is worth b c d. It is squared by filling the voids n m with the segment a.

This one is double - d - o

+ n-cab

a is worth b c d. It is squared by filling the voids n o with the two half-segments b c.

a – n o – d c b

a is worth b c d. It is squared by filling the voids o n with the two half-segments of a.

a – n - b

a is worth b. It is squared by filling the void n with the value of the dotted segment a.

+ - b n-d c-a

a is worth n b, and it is squared by putting the lune n into the void of the segment d.

S b m - o n - f a

a is worth the lune b. It is squared by putting the lune m into o and the S into f in the voids n o. Therefore S is worth f.

p o -r n - S a

a is worth o p S. Take away o p and fill n r with its value, and you have made the +, because o p S is worth n b +.

c a – e n d b

a is worth b c d e. It is squared by filling the ordinary voids, that is, n with b c e, and d n will be equal to a.

d – a – c f e b

a is worth the lunes b c. It is squared by filling the voids e f with the convexities of the 2 lunes b c.

b - a - o n - d c

a is worth b c d. It is squared by filling the voids n o with the segment b.

c - o a - e d n b

a is worth b c d e. It is squared by filling n o with d c, and the concavity left by d is filled by the convexity of b and of c.

b — d— a— o n c

a is worth b c d. It is squared with b divided into n o, and with the convexity of c together with the convexity of d, which fill the concavity of that same d.

n – a - b

a is worth b with its supplements. It is squared with the arrow n that cuts the small rectilinear triangle equal to the lune.

a b - o p - i

a is worth b with its supplements. The whole upper lune goes into o, and the lower lune is worth p, truncated.

o m - a - n

a is worth n m. It is squared by drawing n out of a; the remainder of a is worth m.

a - c m b

a is worth b c. It is squared by filling the void m with the value of the two lunes b c.

o p n – m - a

a is worth the lune n p o. It is squared by filling m with the two segments n o.

+o c + n – d m b - a

a is worth b c d. It is squared by filling the greatest empty segment with the two lunes b d.

c -b a

The dotted half-segments a b fill with themselves the whole empty segment.

a - b n

a is worth b, and it is squared by filling the oval void n with the dotted segment a.

n – b a m

a is worth b with its supplement. It is squared by filling the void m with the dotted segment b.

c b - m n - a

a is worth b c. It is squared by filling the voids n m with the dotted segment c.

b - a - m n

a is worth b. It is squared by filling the voids n m with the dotted segment b.

o - b a

a is worth b. It is squared by filling the empty segment o with the dotted lune.

b – a n

a is worth b. It is squared by filling the void n with the dotted segment b.

+ d - c b - a

This one is duplicated here and is the last in the next-to-last row on the left hand.

a - b

a is worth b with its supplements; which supplements are squared as the seventh of the next-to-last row shows; or else draw b out of a, and the remainder of a is worth the two lunes.

p -i o

i o is worth o p.

e – m n – f c a b d

a b c is worth d e f. It is squared by filling the voids n m with the dotted half-segments d f.

e – f c a d – m b n

a b c is worth d e f. It is squared by filling the voids n m with the two dotted half-segments d f.

b - d a c

a is worth b. It is squared by filling the two voids c d with the dotted segment b.

a — m n — d c b

a is worth b c d. It is squared by filling the voids n m with the dotted segment a.

edcb – m a n

a b c is worth d e. It is squared by filling the two voids n m with the two dotted half-segments d e.

b – m n – a – o c

a is worth b c o. It is squared by filling the two empty segments n m with the two dotted half-segments of b.

S p r – m f – n a o

a is worth o p m. It is squared by filling the voids f m with the two segments r S.

b – n o - a

a is worth b with its supplements. It is squared by putting the two dotted lunes into the segments o n.

d - c a b

a is worth b c d with its supplements. It is squared in the same way as the preceding.

b a - m

Half of this is worth the preceding triangle, and it is squared by putting the value of the two segments a b into m.

n – b c a

a b is worth c with its supplements, which supplements fill the empty triangle n.

+c – b o n a

a b is worth c. It is squared by filling the void n o with the two dotted lunes of c.

It is not squared by itself: b n a – p o

a b is worth n o p with its supplements. It is squared by drawing n out of a b; the remainder of a b is rectilinear and is worth o p with its supplements.

c - b a - d

a is worth b, is worth d; and draw d c out of a b; the remainder of a b will be squared and will be equal in value to the supplements of d c.

c — b a — d

a b is worth c d. It is squared by filling the two smaller empty segments p o with the segment c.

b a - e n o d - c

a b is worth c d e. It is squared by filling the voids o n with the two half-segments d e.

c – d – n o – b a

a b is worth c d. It is squared by filling the two empty segments o n with the segment c.

c — no— ebad

a b is worth c d e. It is squared by filling the voids o n with the half-segments d e.

n a — p o

a n goes into o p, and the whole remains squared.

m n - b c a - e d. a b is worth c d e n m.

p n o – f d c e

c d e f is worth o n p. It is squared by filling the voids n m with the two lunes o p.

d – m n – b a - c

a b is worth c d with its supplement. It is squared by filling the voids n m with the lune-shaped supplements.

tcr-on-ba

a b is worth c r t. It is squared by filling half of each void n o with the lunes r t.

+ m n – b a – g f – r S p o

a b is worth the four lunes o n m r, and is worth the two triangles p S; the void f g is filled with the 4 lunes.

a – m n

a b is worth n m with its supplements, which remain squared by the quantity of the two triangles a b.

pco—ba—St

a b is worth c o p. It is squared by putting the two lunes o p into the two empty segments t S.

d c - b a - o

a b is worth c d o with its supplements. It is squared by putting c d into the smaller empty segments.

rco-ba

a b is worth c with its supplements. It is squared by filling with the two lunes between a c b.

S o - c b - a

a is worth b c. It is squared by filling half the width that the 2 voids o S have with the 2 lunes that lie above b c.

q p - a - n o

a is worth p q. It is squared by putting the lunes p q into the empty segments o n.

+ m - b c a - n f o

a b is worth c o n m. It is squared by filling the greatest segment f with the four lunes o m n.

d c — b a

a b is worth c d. It is squared by filling the four empty segments with the two lunes joined with c d.

c -b a -n o

a b is worth c o n. It is squared by drawing c out of a b; the remainder of a b is worth the remainder of c with o n.

b a - c

a b is worth c with its supplements. It is squared by drawing c out of a b; the remainder of that a b is worth the supplements of c.

ead-po-cb

a is worth b c d e. It is squared by filling the two empty segments o p with the two lunes d c.

enod— a— c b

a is worth b c d e. It is squared by filling the two segments o n with the two lunes d e.

d b a c — e

a b is worth c d e with its supplements. It is squared by filling the four smaller segments with the two larger halves.

c-ba

a b is worth c with its supplements. It is squared...

a—dc—no—b

a b is worth c d. It is squared by filling the two voids around the triangle b with the two lunes above c d.

n

This is squared by removing the 4 small segments and, with their value, filling the large segment.

d —fo —cab

a is worth b c d. Take out the segment of b c and put it into o f.

n - m a

a is worth n m. n is 1/4 of the whole, and so is m.

To whom is given back what was taken, there is no lack.

m - n

n is worth m.

a; b

The surface a is worth the square b, by the axiom that says: if equal things are taken from equal things, the remainder will be equal.

These two demonstrations are of the same nature, because the lunes are worth as much as the segments. Therefore, removing the four segments from the one semicircle and the four lunes from the other, the remainders of the two semicircles will remain equal to one another. The black lunes are worth 8 small segments.

Treatise on continuous quantity.