The elements of Euclid, explained and demonstrated in a new and most easie method with the uses of each proposition in all the parts of the mathematicks / by Claude Francois Milliet D'Chales, a Jesuit ; done out of French, corrected and augmented, and illustrated with nine copper plates, and the effigies of Euclid, by Reeve Williams ...

PROPOSITION I. PROBLEM.

IF Two Right Lines be given, and one of them be Divided into as many parts or Segments as you please, the Rectangle comprehended under the Two whole Right Lines, shall be equal to all the Rectangles contained under the Line which is not Divided, and the several Segments of the divided Line.

Let the Lines proposed be AB, AC; and let AB be Divided into as many parts as you please, the Rectangle AD comprehended under the Lines AB, AC, is equal to the Rectangle AG, comprehended under AC, AE; and to the Rectangle EH comprehended under EG, equal to AC, and under EF; and to the Rectangle FD, com∣prehended under FH equal to AC, and under FB.

Demonstration. The Rectangle AD is equal to all its parts taken together, which are the Rectangles AG, EH, and FD. There being no other Rect∣angle remaining, therefore the Rect∣angle AD is equal to the Rectangles AG, EH, FD, taken together.

Arithmetically or by Numbers.

THe same Proposition is verified by Numbers. Let us suppose that the Line AC is in Length Five Foot, AE two Foot, EF Four, FB three; and by consequence AB Nine, the Rectangle comprehended under AC Five, and AB Nine, that is to say Five times Nine or Forty Five, is equal to twice Five or Ten; and Four times Five or Twenty, and Three times Five or Fifteen for Ten, Twenty, and Fifteen, added together is equal to Forty Five.

USE.

A,		53
B,		8
C,	50	3
B,		8
D,		24
E,		400
F,		424

THis Proposition Demon∣strateth the ordinary practice of Multiplication. For example, let the number A, 53, be to be Multiplyed by the Line AB, represented by the Number B, 8. I cut or seperate the number A, into as many parts as it hath charact∣ers: For Example, into 50 and 3; which I Multiply by 8, saying 8 times 3 is 24; and

by so doing I make a Rectangle. Again I Multiply the Number 50 by 8; the Product shall be E, 400. It is evident, that the Product of 8 times 53, which is 424, is equal to the Product of 24, and to the Product of 400, they being added or taken together.

PROPOSITION II. THEOREM.

THe Square of a Line is equal to all the Rectangles comprehended under the whole, and under its parts.

Let AB be the proposed Line, and its Square ABDC. I say that the Square ABDC is equal to a Rectangle com∣prehended under the whole Line AB, and under AE; and to a Rectangle comprehended under AB, EF; and to a Third comprehended under AB, and FB.

Demonstration. The Square ABDC is equal to all its parts taken together, which are the Rectangles AG, EH, FD. The first of which is comprehended under AC equal to AB, and under AE. The Second EH, is compre∣hended under EG equal to AC or AB,

and under EF. And the Third FD is comprehended under FH, equal to AB, and under FB: And it being the same thing to be comprehended under a Line equal to AB, as to be compre∣hended under AB it self. Therefore the Square of AB is equal to all the Rectangles comprehended under AB, and under AE, EF, FB, parts of AB.

ARITHMETICALLY.

LEt the Line AB represent the Number Nine; its Square shall be 81. Let the part thereof AE be Four; EF, Three; FB, Two: Nine times Four is Thirty Six, Nine times Three is Twenty Seven, and Nine times Two is Eighteen. It is evident that 36, 27, and 18, added together are 81.

USE.

THis Proposition serveth to prove Multiplication; as also for Alge∣briacal Equations.

PROPOSITION III. THEOREM.

IF a Line be Divided into Two parts, the Rectangle comprehended under the whole Line, and under one of its parts, is equal to the Square of the same part, and to the Rectangle comprehended under both its parts.

Let the Line AB be Divided into Two parts in the point C; and let there be made a Rectangle comprehended under AB, and one of its parts; for example AC, that is to say, that AD be equal to AC; and compleat of the Rectangle AF, which shall be equal to the Square of AC, and to the Rectangle comprehended under AC, CB. Draw the Perpendicular CE.

Demonstration. The Rectangle AF, comprehended under AB, and under AD equal to AC, is equal to all its parts, which are the Rectangles AE, CF. The first AE is the Square of AC, seeing that the Lines AC, AD, are equal; and the Rectangle CF is comprehended under CB, and under CE, equal to AD, or AC. Therefore the Rectangle comprehended under

AB, AC, Is equal to the Square of AC, and to a Rectangle comprehended under AC, CB.

ARITHMETICALLY.

LEt AB be 8, AC, 3; CB, 5; the Rectangle comprehended under AB and AC, shall be Three times 8, or 24; the square of AC, 3, is 9; the Rectangle comprehended under AC, 3; and CB 5, is 3 times 5, or 15. It is evident, that 15 and 9 are 24.

USE.

A		43
C	40.	3
B,		3
120.		9.
129

THis Proposition serveth likewise to Demonstrate the ordinary practice of Mul∣tiplication. For Example, if one would Multiply the Number 43 by 3, having se∣parated the Number of 43 into two parts in 40, and 3; three times 43 shall be as much as three times 3, which is Nine, the Square of Three; and Three times Forty, which is 120; for 129 is Three Times 43. Those which are young beginners ought not to be discouraged,

if they do not conceive immediately these Propositions, for they are not difficult, but because they do imagine they contain some great Mystery.

PROPOSITION IV. THEOREM.

IF a Line be Divided into Two Parts, the Square of the whole Line shall be equal to the Two Squares made of its parts, and to Two Rectangles compre∣hended under the same parts.

Let the Line AB be Divided in C, and let the Square thereof ABDE be made; let the Diagonal EB be drawn, and the Perpendicular CF cutting the same; and through that Point let there be drawn GL Parallel to AB. It is evident, that the Square ABDE is equal to the Four Rectangles GF, CL, CG, LF. The Two first are the Square, of AC, and of CB; the Two Complements are comprehended under AC, CB.

Demonstration. The Sides AE, AB, are equal; thence the Angles AEB, ABE, are half Right; and because of the Parallels GL, AB, the Angles of

the Triangles of the Square GE, (by the 29^th,) shall be equal; as also the Sides (by the 6^th of the 1.) Thence GF is the Square of AC. In like manner the Rectangle CL is the Square of CB; the Rectangle GC is comprehended under AC, and AG equal to BL, or BC; the Rect∣angle LF is comprehended under LD equal to AC, and under FD equal to BC.

Coroll. If a Diagonal be drawn in a Square, the Rectangles through which it passeth are Squares.

USE.

A,	144
B,	22
C,	12

THis Proposition giveth us the practical way of finding or extracting the Square Root of a Number propounded. Let the same be the number A, 144, repre∣sented by the Square AD, and its Root by the Line AB. Moreover I know that the Line required AB must have Two Figures. I therefore imagine that the Line AB is Divided in C, and that AC re∣presenteth the first Figure; and BC the Second. I seek the Root of the First Figure of the Number 144, which is 100, and I find that it is 10; and making its Square

100, represented by the Square GF; I Subtract the same from 144, and there remains 44 for the Rectangles GC, FL, and the Square CL. But because this gnomonicall Figure is not proper, I trans∣port the Rectangle FL, in KG, and so I have the Rectangle KL containing 44. I know also almost all the Length of the Side KB; for AC is 10, therefore KC is 20: I must then Divide 44, by 20; that is to say, to find the Divisor, I double the Root found, and I say how many times 20 in 44? I find it 2 times, for the Side BL; but because 20 was not the whole Side KB but only KC; this 2 which cometh in the Quotient is to be added to the Divisor, which then will be 22. So I find the same 2 times precisely in 44, the Square Root then shall be 12. You see that the Square of 144 is equal to the Square of 10, to the Square of 2, which is 4, and to twice 20, which are Two Rect∣angles comprehended under 2, and under 10.

PROPOSITION V. THEOREM.

IF a Right Line be cut into equal parts, and into unequal parts; the Rectangle comprehended under the unequal parts, to∣gether with the Square which is of the middle part, or difference of the parts, is equal to the Square of half the Line.

If the Line AB is Divided equally in C, and unequally in D; the Rect∣angle AH comprehended under the Segments AD, DB, together with the Square of CD, shall be equal to the Square CF; that is of half of AB, viz. CB. Make an end of the Figure as you see it; the Rectangles LG, DI, shall be Squares (by the Coroll. of the 4^th.) I prove that the Rectangle AH, com∣prehended under AD, and DH, equal to DB, with the Square LG, is equal to the Square CF.

Demonstration. The Rectangle AL, is equal to the Rectangle DF; the one and the other being comprehended under half the Line AB, and under BD, or DH equal thereto. Add to both the Rectangle CH; the Rectangle

AH, shall be equal to the Gnomon LBG. Again to both add the Square LG; the Rectangle AH, with the Square LG, shall be equal to the Square CF.

ARITHMETICALLY.

LEt AB be 10; AC is 5, as also CB. Let CD be 2, and DB, 3; the Rectangle comprehended under AD, 7; and DB, 3: that is to say 21, with the Square of CD 2, which is 4, shall be equal to the Square of CB 5, which is 25.

USE.

THis Proposition is very useful in the Third Book; we make use thereof in Algebra, to Demonstrate the way of finding the Root of an affected Square or Equation.

PROPOSITION VI. THEOREM.

IF one add a Line to another which is Divided into Two equal parts; the Rectangle comprehended under the Line compounded of both, and under the Line added, together with the Square of half the Divided Line, is equal to the Square of a Line compounded of half the Divided Line, and the Line added.

If one add the Line BD, to the Line AB, which is equally Divided in C; the Rectangle AN, comprehended under AD, and under DN, or DB, with the Square of CB, is equal to the Square of CD. Make the Square of CD, and having drawn the Diagonal FD, draw BG Parallel to FC, which cuts FD in the Point H, through which passeth HN Parallel to AB: KG shall be the Square of BC; and BN, that of BD.

Demonstration. The Rectangles AK, CH, on equal Bases, AC, BC, are equal (by the 38^th of the 1^st.) The Com∣plements CH, HE, are equal (by the 43^d of the 1^st.) Therefore the Rectangles

AK, HE, are equal. Add to both the Rectangle CN, and the Square KG, the Rectangles AK, CN, that is to say the Rectangle AN with the Square KG, shall be equal to the Rect∣angles CN, HE, and to the Square KG, that is to say to the Square CE.

Arithmetically or by Numbers.

LEt AB be 8, AC 4, CB 4, BD 3; then AD shall be 11. It is evi∣dent that the Rectangle AN, three times 11, that is to say 33 with the Square of KG 16, which together are 49, is equal to the Square of CD 7, which is 49, for 7 times 7 is 49.

USE.

MAurolycus measured the whole Earth by one single Observation, making use of this Proposition. He would that one should observe from A the top of a Mountain whose height is known; the Angle BAC, which is made by the Line AB which toucheth the superficies of the Earth, and the Line AC which passeth through the Center; and that in the Tri∣angle ADF; knowing the Angle A,

and the Right Angle ADF, one findeth by Trigonometry the Sides AF, FD; and because that it is easie to Demonstrate that FD, FB, are equal, one shall know the Length of the Line AB, and its Square. Now having Demonstrated by the preceding Proposition, that the Line ED being Divided into two equal parts in the Point C, and having added thereto AD; the Rectangle comprehended under EA, and under AD, together with the Square of CD, or CB, is equal to the Square of AC, and the Angle ABC being Right, (as it is proved in the Third Book) the Square of AC is equal to the Squares of AB, BC. Therefore the Rectangles AE, AD, together with the Square of BC, is equal to the Squares of AB, BC. Take from both the Square of BC; the Rectangle under AE, AD, shall be equal to the Square of AB. Di∣vide then the Square of AB, which you know by the Height of the Mountain, which is AD; the Quotient shall be the Line AE, from which must be Sub∣tracted the Height of the same Mountain; and you shall have DE, the Diameter of the Earth.

We make use of the same Proposition in Algebra; as also to Demonstrate the

practice of finding the Root of a Square equal to a Number encreased more by a certain number of Roots. The Two Pro∣positions following serveth also to prove other like Practises.

PROPOSITION VII. THEOREM.

IF one Divide a Line, the Square of the whole Line, together with that of one of its parts, shall be equal to Two Rectangles comprehended under the whole Line, and that first part, and to the Square of the other part.

Let the Line AB be Divided at discretion, in the Point C; the Square AD of the Line AB, together with the Square AL, shall be equal to Two Rectangles comprehended under AB, AC. With the Square of CB make the Square AB; and having drawn the Diagonal EB, and the Lines CF, HGI, prolong EA, and make AK equal to AC; so shall AL be the Square of AC, and HK shall be equal to AB. For HA is equal to GC; and GC is equal to CB, seeing CI is the Square of CB, (by the Coroll. of the 4^th.)

Demonstration. It is evident that the Squares AD, AL, are equal to the Rectangles HL, HD, and to the Square of CI. For the Rectangle HL is comprehended under HK equal to AB, and under KL equal to AC. In like manner, the Rectangle HD is comprehended under HI, equal to AB, and under HE equal to AC. Therefore the Squares of AB, AC, are equal to Two Rectangles comprehended under AB, AC, and to the Square of CB.

ARITHMETICALLY.

LEt it be supposed that the same AB contains 9 equal parts; AC, 4; CB, 5: the Square of AB, 9, is 81; the Square of AC 4, is 16. Which being added together is 97. A Rectangle under AB, AC, or 4 times 9 is 36; which doubled is 72; The Square of CB, 5, is 25. Now 25 and 72 is also 97.

PROPOSITION VIII. THEOREM.

IF one Divide a Line, and that thereto one add one of its parts, the Square of the Line Composed shall be equal to Four Rectangles, comprehended under the First Line, and under the part added together with the Square of the other part.

Let the Line AB be Divided at dis∣cretion in the Point C; and let thereto be added BD, equal to CB, the Square of AD shall be equal to 4 Rectangles comprehended under AB, BC or BD, and to the Square of AC. Let the Square of AD be made, and having drawn the Diagonal AE, draw the Perpendiculars BP, CN, which cut the Diagonal in L and O; draw also the Lines MIH, GLR, Parallel to AB; the Rectangles GC, LK, PH, MB, NR, shall be Squares (by the Coroll. of the 4^th.)

Demonstration. The Square ADEF is equal to all its parts; the Rectangles IB, OD, PM, are comprehended under Lines equal to AB and CB. If you add the Rectangle ML to the Rect∣angle PH; you shall have a Rectangle

comprehennded under one Line equal to AB, and under another equal to CB or BD. There remains only the Square GC, which is that of AC. Therefore the Square of AD is equal to Four Rectangles comprehended under AB, BD, and to the Square of AC.

ARITHMETICALLY.

LEt the Line AB contain 7 parts; AC 3, CB 4, as well as BD; the Square of AD, 11, shall be 121. A Rectangle under AB, 7, and BD, 4; is Twenty Eight: which taken 4 times, is 112. and the Square of 3 is 9. Now 112. and 9. added are 121.

PROPOSITION IX. THEOREM.

IF a Line be equally Divided, and un∣equally; the Squares of the unequal parts shall be double to the Squares of half the Line, and of the part between.

Let the Line AB be Divided equally, in the Point C; and unequally, in the Point D. The Squares of the unequal parts AD, DB, shall be double of the

Squares of AC, which is the half of AB, and the Square between both CD. Draw AB, and the Perpendi∣cular CE, equal to AC: Draw also the Lines AE, BE, and the Perpendi∣cular DF; as also FG Parallel to CD. Then draw the Line AF.

Demonstration. The Lines AC, CE, are equal, and the Angle C is Right; thence (by the 6^th of the 1^st.) the Angles CAE, CEA, are equal to a half Right Angle. In like manner, the Angles CEB, CBE, CFE, DFB, are half Right; the Lines GF, GE, DF, DB, are equal; and the total Angle AEF is Right. The Square of AE (by the 47^th of 1^st.) is equal to the Squares of AC, CE, which are equal: Thence it is double to the Square of AC. After the same manner the Square of EF is double to the Square of GF, or CD: Now the Square of AF is equal to the Squares of AE, EF; seeing that the Angle AEF is Right: There∣fore the Square of AF is double to the Squares of AC, CD. The same Square AF is equal to the Squares of AD, DF, or DB; seeing that the Angle D is Right: There∣fore the Squares of AD, DB, are

double to the Squares of AC, CD.

ARITHMETICALLY.

LEt AB be 10; AC, 5; CD, 3; DB, 2: The Squares of AD, 8, and DB, 2; that is to say 64, and 4, which together are 68, are double to the Square AC, 5, which is 25; and of the Square CD, 3, which is 9; for 25 and 9 are 34, which doubled make 68.

USE.

I Have not found this Proposition, nor the insuing, but only in Algebra.

PROPOSITION X. THEOREM.

IF one add a Line, to another which is equally Divided; the Square of the Line Composed of both, with the Square of the Line added, are double to the Square of half the Line, and to the Square which is Composed of the half Line and the Line added.

If one supposeth AB, to be Divided in the middle at the Point C; and if thereto be added the Line BD, the Squares of AB and BD, shall be double to the Squares of AC, and CD, added together. Draw the Perpendiculars CE, DF, equal to AC: Then draw the Lines AE, EF, AG, EBG.

Demonstration. The Lines AC, CE, CB, beng equal, and the Angles at the Point C being Right: The Angles AEC, CEB, CBE, DBG, DGB, shall be half Right; and the Lines DB, DG, and EF, FG, CD, shall be equal. The Square of AE, is double to the Square of AC; the Square of EG, is double to the Square of EF, or CD, (by the 47^th of the 1^st.) Now the Square of AG, is equal to the Squares of AE, EG, (by the 47^th of the 1^st:) Therefore the Square of AG is double to the Squares of AC, CD. The same AG (by the 47^th of the 1^st,) is equal to the Squares of AD, BD, or GD: Therefore the Squares of AD, BD, are double the Squares of AC, CD.

ARITHMETICALLY.

LEt AB be 6 parts, AC, 3; CB, 3; BD, 4: the Square of AD, 10, is 100. the Square of BD, 4, is 16, which are 116. The Square of AC, 3, is 9; the Square of CD, 7, is 49. Now 49. and 9, is 58, the half of 116.

PROPOSITION XI. PROBLEM.

TO Divide a Line, so that the Rect∣angle comprehended under the whole Line, and under one of its parts, shall be equal to the Square of the other part.

It is proposed to Divide the Line AB, so that the Rectangle comprehended under the whole Line AB, and under HB, be equal to the Square of AH. Make a Square of AB (by the 46^th of the 1^st:) Divide AD in the middle in E; then Draw EB, and make EF equal to EB. Make the Square AF, that is to say that AF, AH, be equal. I say that the Square of AH, shall be equal to the Rectangle HC, compre∣hended

under HB, and the Line BC, equal to AB.

Demonstration. The Line AD is equally divided in E, and there is ad∣ded thereto the Line FA; thence (by the 6^th.) the Rectangle DG, compre∣hended under DF and FG, equal to AF, with the Square of AE, is equal to the Square of EF, equal to EB. Now the Square of EB is equal to the Squares of AB, AE, (by the 47^th of the 1^st.) therefore the Squares of AB, AE, are equal to the Rectangle DG, and to the Square of AE: and taking away from both, the Square of AE, the Square of AB, which is AC, shall be equal to the Rectangle DG: taking also away the Rectangle DH, which is in both, the Rectangle HC shall be equal to the Square of AG.

USE.

THis Proposition serveth to cut a Line in extream and mean Proportion; as shall be shewn in the Sixth Book. It is used often in the 14^th. of Euclid's Ele∣ments, to find the Sides of Regular Bodies. It serveth for the 10^th. of the Fourth Book, to inscribe a Pentagone in a Circle, as al∣so

a Pentadecagone. You shall see other uses of a Line thus divided, in (the 30^th. of the Sixth Book)

PROPOSITION XII. THEOREM.

IN an obtuse angled Triangle the Square of the side opposite to the obtuse Angle, is equal to the Squares of the other two sides, and to two Rectangles comprehended under the side on which one draweth a Perpendicular, and under the Line which is between the Triangle, and that perpen∣dicular

Let the Angle ACB of the Triangle ABC, be obtuse; and let AD be drawn perpendicular to BC. The Square of the side AB, is equal to the Squares of the sides AC, CB, and to two Rectangles comprehended under the side BC, and under DC.

Demonstration. The Square of AB is equal to the Squares of AD, DB. (by the 47^th. of the 1^st.) the Square of DB is equal to the Squares of DC and CB, and to two Rectangles compre∣hended under DC, CB, (by the 4^th.) therefore the Square of AB is equal to

the Squares of AD, DC, CB, and to two Rectangles comprehended under DC, CB, in the place of the two last Squares AD, DC. Put the Square of AC, which is equal to them (by the 47^th of the 1^st.) the Square of AB shall be equal to the Square of AC and CB. and to two Rectangles comprehended under DC, CB.

USE.

THis Proposition is useful to measure the Area of a Triangle, its three sides being known: for Example; If the side AB was twenty Foot; AC, 13; BC, 11; the Square of AB would be four hundred; the Square of AC, one hundred sixty nine; and the Square of BC, one hundred twenty one: the Sum of the two last is, Two hundred and nine∣ty; which being subtracted from four hundred, leaves one hundred and ten for the two Rectangles under BC, CD, the one half-fifty five shall be one of those Rectangles: which divided by BC 11, we shall have five for the Line CD, whose Square is twenty five; which being subtracted from the Square of AC, one hundred sixty nine, there remains the Square of AD, one

hundred forty four; and its Root shall be the side AD: which being multiplied by 5½, the half of BC, you have the Area of the Triangle ABC, containing 66 square Feet.

PROPOSITION XIII. THEOREM.

IN any Triangle whatever, the Square of the side opposite to an acute Angle, together with two Rectangles comprehend∣ed under the side on which the Perpendi∣cular falleth, and under the Line which is betwixt the Perpendicular and that An∣gle, is equal to the Square of the other sides.

Let the proposed Triangle be ABC, which hath the Angle C, acute; and if one draw AD perpendicular to BC, the Square of the side AB, which is opposite to the acute Angle C, toge∣ther with two Rectangles comprehend∣ed under BC, DC, shall be equal to the Squares of AC, BC.

Demonstration. The Line BC is di∣vided in D; whence (by the 7^th.) the Square of BC, DC, are equal to two Rectangles under BC, DC, and to

the Square of BD: add to both the Square of AD; the Square of BD, DC, AD, shall be equal to two Re∣ctangles under BC, DC; and to the Squares of BD, AD, in the place of the Squares of CD, AD, put the Square of AC, which is equal to them (by the 47^th. of the 1^st.) and instead of the Squares of BD, AB, substitute the Square of AB, which is equal to them, the Squares of BC, AC, shall be equal to the Square of AB, and to two Rect∣angles comprehended under BC, DC.

USE.

THese Propositions are very necessary in Trigonometry: I make use there∣of in the eighth Proposition of the third Book, to prove, That in a Triangle there is the same Reason between the whole Sine and the Sine of an Angle, as are between the Rectangle of the sides comprehending that Angle, and the double Area of the Triangle. I make use thereof in several other Propositions, as in the seventh.

PROPOSITION XIV. PROBLEM.

TO describe a Square equal to a right lined Figure given.

To make a Square equal to a Right lined Figure A, make (by the 45^th. of the 1^st.) a Rectangle BC, DE, equal to the Right lined Figure A: if its sides CD, DC, were equal, we should have already our desire; if they be un∣equal, continue the Line BC, until CF be equal to CD; and dividing the Line BF in the middle, in the Point G describe the Semicircle FHB, then continue DC to H, the Square of the Line CH is equal to the Right lined Figure A: draw the Line GH.

Demonstration. The Line BF is equally divided in G, and unequally in C; thence (by the 5^th.) the Rectangle comprehended under BC, CF, or CD, that is to say, the Rectangle BD, with the Square CG, is equal to the Square of GB, or to its Equal GH. Now (by the 47^th. of the 1^st.) the Square of GH is equal to the Square of CH, CG, therefore the Rectangle BD, and the

Square of CG, is equal to the Squares of CG, and of CH; and taking away the Square of CG, which is common to both, the Rectangle BD, or the Right lined Figure A, is equal to the Square of CH.

USE.

THis Proposition serveth in the first place, to reduce into a Square any Right lined Figure whatever: and whereas a Square is the first Measure of all Super∣ficies, because its Length and Breadth is equal, we measure by this means all right lined Figures. In the second place this Proposition teacheth us to find a mean Pro∣portion between two given Lines; as we shall see in the Thirteenth Proposition of the Sixth Book.

This Proposition may also serve to square curve lined Figures, and even Circles themselves; for any crooked or curve lined Figure may, to sence, be reduced to a Right lined Figure; as if we inscribe in a Circle a Polygon having a thousand sides, it shall not be sensibly different from a Circle: and reducing the Polygone into a Square, we square nearly the Circle.

Notes

• Fig. 6.