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.

To find the Center of a Circle.

IF you would find the Center of the Circle AEBD, draw the Line AB, and divide the same in the middle, in the Point C; at which Point erect a Perpendicular ED, which you shall di∣vide also equally in the Point F. This Point F shall be the Center of the Circle; for if it be not, imagine, if you please,

that the Point G is the Center: draw the Lines GA, GB, GC.

Demonstration. If the Point G were the Center, the Triangles GAC, GBC, would have the sides GA, GB, equal by the definition of a Circle: AC, CB, are equal to the Line AB, having been divided in the middle, in the Point C. And CG being common, the Angles GCB, GCA, would then be equal (by the 8^th. of the 1^st.) and CG would be then a Perpendicular, and not CD; which would be contrary to the Hy∣pothesis. Therefore the Center cannot be out of the Line CD. I further add, that it must be in the Point F, which divideth the same into two equal Parts; otherwise the Lines drawn from the Center to the Circumference would not be equal.

Corollary. The Center of a Circle is in a Line which divideth another Line in the middle, and that perpendicularly.

USE.

THis first Proposition is necessary to demonstrate those which follow.

PROPOSITION II. THEOREM.

A Streight Line drawn from one point of the Circumference of a Circle to another, shall fall within the same.

Let there be drawn from the Point B, in the Circumference, a Line to the Point C. I say that it shall fall whol∣ly within the Circle. To prove that it cannot fall without the Circle, as BVC; having found the Center there∣of, which is A, draw the Lines AB, AC, AV.

Demonstration. The Sides AB, AC, of the Triangle ABC, are equal: whence (by the 5^th. of the 1^st.) the An∣gles ABC, ACB, are equal. And seeing the Angle AVC, is exteriour in respect of the Triangle AVB, it is greater than ABC, (by the 16^th. of the 1^st.) it shall be also greater than the Angle ACB. Thence (by the 19^th of the 1^st.) in the Triangle ACV, the side AC, opposite to the greatest An∣gle AVC, is greater than AV; and by consequence, AV cannot reach the circumference of the Circle, seeing it

is shorter than AC, which doth but reach the same; wherefore the Point V is within the Circle: the same may be proved of any Point in the Line AB; and therefore the whole Line AB falls within the Circle.

USE.

IT is on this Proposition that are ground∣ed those which demonstrate that a Circle toucheth a streight Line but only in one Point: for if the Line should touch two Points of its Circumference, it would be then drawn from one Point of its Cir∣cumference to the other; and consequent∣ly, would fall within the Circle, accord∣ing to this Proposition; although its De∣finition saith, that it cutteth not its Cir∣cumference. Theodosius demonstrateth after the same manner, That a Globe can∣not touch a Plain but in one single Point; otherwise the Plain would enter into the Globe.

PROPOSITION III. THEOREM.

IF the Diameter of a Circle cutteth a Line which passeth not through the Center, into two equal Parts, it shall cut the same at right Angles; and if it cut∣teth it at Right Angles it divideth it into two equal Parts.

If the Diameter AC cut the Line BD, which passeth not through the Center F, into two equal Parts in E, it shall cut the same at Right Angles. Draw the Lines FB, FD.

Demonstration. In the Triangles FEB, FED, the side FE is common; the sides BE, ED, are equal, seeing the Line BD is divided equally in the Point E; the Bases FB, FD, are equal: thence (by the 8^th. of the 1^st.) the Angles BEF, DEF, are equal, and consequently right. Moreover, I say, that if the Angles BEF, DEF, are right, the Line BD shall be divided into two equal Parts in E; that is to say, the Lines BE, ED, shall be equal.

Demonstration. The Triangles BEF, DEF, are Rectangular: thence (by the 47^th. of the 1^st.) the Square of the side FD shall be equal to the Squares of ED, and of EF: in like manner, the Square of BF shall be equal to the Squares of BE, EF. Now the Squares of BF, FD, are equal, because the Lines are equal: therefore the Squares of BF, EF, are equal to the Squares of ED, EF; and taking away EF, the Squares of BE, ED, shall be equal, and conse∣quently the Lines.

PROPOSITION IV. THEOREM.

TWo Lines drawn in a Circle, cut not each other equally, if the Point of their Intersection be not in the Center.

If the Lines AC, BD, cut each other in the Point I, which is not the Cen∣ter of the Circle, they will not cut each other equally. In the first place, if one of them should pass through the Center, it is evident, that the same cannot be divided into two equal Parts, but by the Center. If neither of them pass through the Center, as BD, EI, draw the Line AIC, which passeth through the Center.

Demonstration. If the Line AC di∣vide the Line BD into two equal Parts in I, the Angles AID, AIB, would be equal (by the 2^d.) In like manner, if the Line EG was divided into two equal Parts, in the Point I, the Angle AIG would be right: thence the Angles AIB, AIG, would be right, and con∣sequently equal, which cannot be, the one being part of the other. Lastly, the Line AIC, which passeth through the Centers, would then be perpendi∣cular to the Lines BD, GI, if both of them were divided equally in the Point I.

USE.

BOth of those Propositions are of use in Trigonometry: hereby is demonstra∣ted, that the half Chord of an Arch is per∣pendicular to the Semidiameter; and by consequence, it is the sine of the half Arch: by which also it is demonstrable, that the Side of a Triangle hath the same Propor∣tion as the sines of their opposite Angles. We furthermore make use of this Proposi∣tion to find the Excentricity of the Circle which the Sun describeth in a Year.

PROPOSITION V. THEOREM.

CIrcles which cut each other have not the same Center.

The Circles ABC, ADC, which cut each other in the Points A and C, have not the same Center: for if they had the same Center E, the Lines EA, ED, would then be equal (by the de∣finition of a Circle,) as also the Lines EA, EB: thence the Lines ED, EB, would be equal; which is impossible, the one being a part of the other.

PROPOSITION VI. THEOREM.

IF two Circles inwardly touch one the other, they have not one and the same Center.

The Circles BD, BC, which touch inwardly one the other, in the Point B, have not the same Center: for if the Point A was the Center of both Circles, the Lines AB, AD, and AB, AC, would be equal (by the definition of a

Circle) and so AD, AC, would be equal; which is impossible, the one being a Part of the other.

PROPOSITION VII. THEOREM.

IF in a Circle several Lines be drawn from a Point which is not the Center, unto the Circumference; first, that which passeth through the Center shall be great∣est; secondly, the least shall be the Re∣mainder of the same Line; thirdly, the nearest Line to the greatest is greater than any Line which is farther from it; fourthly, there cannot be drawn more than two equal Lines.

Let there be drawn several Lines from the Point A, which is not the Center of the Circle, to the Circumfe∣rence; and let the Line AC pass through the Center B: I demonstrate that it shall be the greatest or longest, that it is greater than AF. Draw FB.

Demonstration. The sides AB, BF, of the Triangle ABF, are greater than the side AF, (by the 20^th. of the 1^st.) Now BF, BC, are equal (by the defini∣tion of a Circle) therefore AB, BC,

that is to say, AC, is greater than AF. Moreover, in the second place, I say, that AD is the shortest; for Example, shorter than AE. Draw EB.

Demonstration. The Sides EA, AB, of the Triangle ABE, are greater than BE, which is equal to BD; thence EA, AB, are greater than BD; and taking away AB, which is common, AE shall be greater than AD. More∣over, AF, which is nearer to AC than AE, is greater than AE.

Demonstration. The Triangles FBA, EBA, have the Sides BF, BE, equal, and BA common to both: the Angle ABF is greater than the Angle ABE; thence (by the 24^th. of the first) AF is greater than AE.

In fine, I say that there cannot be drawn but two equal Lines from the Point A, to the Circumference, let the Angles ABE, ABG, be equal, and let be drawn the Line AE, AG.

Demonstration. The Triangles ABG, ABE, have their Bases AE, AG, equal; and all the Lines which may be drawn on either side of these, shall be either nearer to AC than the Lines AE, AG, or farther: and so they shall be either shorter or longer than

AE, AG. Therefore, there cannot be drawn more than two equal Lines.

USE.

THeodosius doth very well make use of this Proposition, to prove, that if from a Point in the Superficies of a Sphere, which is not the Pole of a Circle, there be drawn several Arks of great Cir∣cles unto the Circumference of a Circle: that which passeth through the Pole of that Circle to which the great Circles are drawn, is greatest. For Example, if from the Pole of the World, which is not the Pole of the Horizon (for the Zenith is the Pole thereof) there be drawn several Arks of great Circles unto the Circumfe∣rence of the Horizon: the greatest Ark of all shall be that part of the Meridian which passeth through the Zenith. This Proposition is also made use of, to prove, that the Sun being in Apoge, is farthest distant from the Earth.

PROPOSITION VIII. THEOREM.

IF from a Point without a Circle, be drawn several Lines to its Circum∣ference; first, of all those which are drawn ••o its concave Circumference, the greatest ••asseth through the Center; secondly, those ••earest thereto are greater than those ••hich are farther off; thirdly, amongst ••he Lines which fall on the convex Cir∣••umference, the least being continued, ••asseth through the Center; fourthly, the ••earest thereto are least; fifthly, there ••annot be drawn but only two equal, whe∣••her they be drawn to the convex Cir∣••umference, or that they fall on the con∣••ave.

Let there be drawn from the Point A, several Lines to the Circumference ••f the Circle GC, DE. In the first ••lace the Line AC, which passeth ••hrough the Center B, is the greatest ••f all those which fall on the concave Circumference: for Example, it is great∣••r than AD. Draw the Line BD.

Demonstration. In the Triangle ABD ••he Sides AB, BD, are greater than the

Side AD; now the Sides AB, BC, are equal to AB, BD; thence AB, BC, or AC, is greater than AD.

2. AD is greater than AE.

Demonstration. The Triangles ABD, ABE, have the Side AB common, and the Sides BE, BD, equal; and the Angle ABD, is greater than the Angle ABE: thence (by the 24^th. of the 1^st.) the Base AD is greater than the Base AE.

3. AF, which being prolonged pas∣seth through the Center, is the least of those Lines which are drawn to the con¦vex Circumference LFIK. For exam¦ple, it is less than AI. Draw IB.

Demonstration. The Sides AI, IB, are greater than AB, (by the 20^th. of the 1^st.) wherefore taking away the equal Lines BI, BF, AF shall be less than AI.

4. AI is less than AK. Draw the Line BK

Demonstration. In the Triangle•• AIB, AKB, the sides AK, KB, ar•• greater than the Sides AI, IB, (by th•• 21^th. of the 1^st.) wherefore taking away the equal sides, BK, BI, there remain•• AI less than AK.

5. There can be drawn but two equal Lines; make the Angles ABL, ABK, equal.

Demonstration. The Triangles ABL, ABK, shall have their Bases AL, AK, equal (by the 4^th. of the 1^st.) but there cannot be drawn any other, which will not be either nearer or farther from AF, and which will not be either great∣er or lesser than AK, AL.

PROPOSITION IX. THEOREM.

THe Point from whence may be drawn to the Circumference of a Circle, three equal Lines, is its Center.

If that Point were not the Center of the Circle, there could be drawn therefrom only two equal Lines (by the 7^th.)

PROPOSITION X. THEOREM.

TWo Circles cut each other only in two Points.

If the two Circles ABD, ABFD, did cut each other in three Points, A,

B, D; seek (by the 2^d.) the Center C of the Circles AE, BD, and draw the Lines CA, CB, CD.

Demonstration. The Lines AC, BC, DC, drawn from the Center C, to the Circumference of the Circle AE, BD, are equal: Now the same Lines are also drawn to the Circumference of the Circle AB, FB: thence (by the 9^th.) the Point C shall be the Center of the Circle ABFD. So two Circles which cut each other shall have the same Center; which is contrary to the fifth Proposition.

PROPOSITION XI. THEOREM.

IF two Circles inwardly touch one the other, the Line drawn through both Centers, passeth also through the Point, where they touch one the other.

If the two Circles EAB, EFG, in∣wardly touch one the other, in the Point E, the Line drawn through both Centers shall pass through the Point E. Now if the Point D be the Center of the lesser Circle, and C that of the greater, in such manner, that the

Line CD passeth not through the Point E, draw the Lines CE, DE.

Demonstration. The Lines DE, DG, drawn from the Center D of the lesser Circle, are equal; and adding the Line DC; the Lines ED, DC, should be equal to CG. Now ED, DC, are greater than EC (by the 20^th of the 1^st.) so that CG is greater than CE, and notwithstanding C being the Center of the great Circle; CE, CB, are equal; thence CG would be greater than CB, which is impossible.

PROPOSITION XII. THEOREM.

IF Two Circles touch one the other out∣wardly, the Line drawn through their Centers passeth through the Point where they touch one the other.

If the Line AB drawn from the Center A, to the Center B, passeth not through the Point C, where the Circles touch one the other; draw the Lines AC, BC.

Demonstration. In the Triangle ACB, the Side AB would be greater than AC, BC; for BC, BE, are equal, as well as AD, AC; which would be contrary to the 20th Proposition of the 1st.

PROPOSITION XIII. THEOREM.

TWo Circles touch one the other only in one Point.

In the first place, if Two Circles touch one the other inwardly, they shall touch one the other only in one Point C, in the Line BAC, which passeth through their Centers A and B. Now if it be supposed that they touch one the other again in the Point D; draw the Lines AD, DB.

Demonstration. The Lines AD, AC, being drawn from the Center of the lesser Circle, are equal, and adding to both AB the Lines BAAC, BA, AD should be equal: Now BC, BD being drawn from the Center of the greater Circle, should be also equal; thence the Sides BA, AD, should be equal to the Side BD;

which is contrary to the 20^th of the 1^st.

Secondly, if the Two Circles touch outwardly; drawing the Line AB from one Center to the other, it shall pass through the Point C, where the Circles touch (by the 12^th.) Now if you say they also touch in the Point D; having drawn the Lines AD, BD; the Lines BD, BC; AD, AC, being equal, the Two Sides of a Triangle taken toge∣ther, should then be equal to the Third, which is contrary to the 20^th of the 1^st.

USE.

THese Four Propositions are very clear, they are notwithstanding necessary in Astronomy, when we make use of Epicy∣cles to explain the motion of the Planets.

PROPOSITION XIV. THEOREM.

EQual Right Lines drawn in a Circle, are equally distant from the Center and Right Lines, which are equally distant from the Center, are equal.

If the Lines AB, CD, are equal; I

demonstrate that the Perpendicular EF, EG, drawn from the Center are also equal. Draw the Lines EA, EC.

Demonstration. The Perpendiculars EF, EG, divideth the Lines AB, CD, in the middle in F and G (by the 3^d:) So AF, CG, are equal, the Angles F and G are Right; thence (by the 47^th of the 1^st) the Square of EA is equal to the Squares of EF, FA; as the Square of EC is equal to the Squares of CG, EG: Now the Squares of EA, EC, are equal, the Lines EA, EC, being equal: Therefore the Squares of EF, FA, are equal to the Squares of EG, GC; and taking away the equal Squares AF, CG; there re∣mains the equal Squares EG, EF; so then the Lines EG, EF, which are the distances are equal.

If you suppose that the distances, or perpendiculars EG, EF, are equal; I would demonstrate after the same manner that the Squares of EF, FA, are equal to the Squares of EG, GC, and taking away the equal Squares of EF, EG, there shall remain the equal Squares AF, CG; so then the Lines AF, CG, and their double AB, CD, are equal.

PROPOSITION XV. THEOREM.

THe Diameter is the greatest Line drawn in a Circle, and those nearest the Center are greatest.

The Diameter AB is the greatest of the Lines that are drawn in the Circle; for example, it is greater than CD. Draw the Lines EC, ED.

Demonstration. In the Triangle CED, the Two Sides EC, ED, are greater than the Side CD (by the 20^th of the 1^st.) Now AE, EB, or AB is equal to EC, ED: Thence the Dia∣meter AB is greater than CD.

Secondly, Let the Line GI, be far∣ther distant from the Center than the Line CD; that is to say, let the perpen∣dicular EH be greater than the Per∣pendicular EF: I say that CD is grea∣ter than GI. Draw the Lines EC, EG.

Demonstration. The Squares CF, FE, (by the 47^th of the 1^st.) are equal to the Square of CE: Now CE is equal to EG, and the Square of EG is equal to the Squares of GH, HE:

By consequence the Squares of CF, FE, are equal to the Squares of GH, HE; and taking away on the one Side the Square of HE, and on the other the Square of FE, which is less; the the Square of CF shall be greater than the Square of GH. Thence the Line CF shall be greater than GH; and the whole Line CD, the double of CF shall be greater than GI, the double of GH.

USE.

THeodosius maketh use of those Two last Propositions to Demonstrate, that in the Sphere, the least Circles are farthest distant from the Center. We make use thereof in Astrolabes. Moreover to those Propositions might be brought the question of Mechanicks, proposed by Aristotle, which certifies that the Rowers which are in the middle of a Galley, has a greater strength than those which are a stern or a head, because that the sides of the Galley being rounded, have the Oars which be there placed longest. Also the De∣monstration of the Rainbow in the Heavens supposeth this Proposition.

PROPOSITION XVI. THEOREM.

THe Perpendicular Line drawn to the extream part of the Diameter is wholly without the Circle, and toucheth the same. All other Lines drawn between it, and the Circumference of the Circle cut it, and go within the same.

Draw by the Point A, which is the extremity of the Diameter AB, the Perpendicular AC: I say first, that all other Points of the same Line, as the Point C, are without the Circle. Draw the Line DC.

Demonstration. Because the Angle DAC of the Triangle DAC, is a Right Angle; DCA shall be acute, and (by the 19^th of the 1^st,) the Side DC shall be greater than the Side DA; thence the Line DC passeth beyond the Circumference of the Circle.

I farther add that the Line CA toucheth the Circle, because in the meeting thereof in the Point A, it cutteth it not, but hath all its other Points without the Circle.

I say also that there cannot be drawn any other Line from the Point A, under∣neath CA, which shall not cut the Circle; as for example, the Line EA. Draw the Perpendicular DI.

Demonstration. Seeing the Angle IAB is Acute, the Perpendicular drawn from the Point D shall be on the same Side with the Angle IAB (by the 17^th of the 1^st.) Let it be DI; the Angle DIA is Right, and the Angle IAD Acute; AD shall be greater than DI; thence the Line DI, doth not reach the Circumference, and the Point I is within the Circle.

PROPOSITION XVII. PROBLEM.

FRom a Point taken without the Circle, to draw a Line to touch the same.

To draw a touch Line, from the Point A to the Circle BD; draw the Line AC to its Center; and at the Point B, the Perpendicular BE; which cutteth in E a Circle described on the Center C, passing through the Point A. Draw also the Line CE. I say that the

Line AD toucheth the Circle in the Point D.

Demonstration. The Triangle EBC, ADC, have the same Angle C; and the Sides CD, CB; CACE, equal, by the Definition of a Circle: so they are equal in every respect (by the 4^th of the 1^st:) and the Angles CBE, CDA are equal: Now CBA is Right; and (by the 16^th,) the Line AD shall touch the Circle.

PROPOSITION XVIII. THEOREM.

THe Line drawn from the Center of a Circle to the Point where a streight Line toucheth the same, is Perpendicular to the same Line.

If there be drawn the Line CD, from the Center C, to the Point of touching D: CD shall be Perpendicular to AB. For if it be not; draw BC Perpendicular to AB.

Demonstration. Since we would have it that the Line CB be Perpen∣dicular; the Angle B shall be Right, and CDB Acute, (by the 32^d of the 1^st.) Thence the Line CB, opposite

to the lesser Angle, shall be lesser than CD; which is impossible, seeing CF is equal to CD.

PROPOSITION XIX. THEOREM.

THe Perpendicular drawn to a touch Line at the Point of touching Passeth through the Center.

Let the Line AB touch the Circle in the Point D; and let the Line DC be Perpendicular to AB: I say that DC passeth through the Center. For if it passeth not through it, drawing from the Center to the Point D a Line; it would be Perpendicular to AB: and so Two Lines Perpendicular to the same, would be drawn to the same Point D; which cannot be.

USE.

THe use of touch Lines is very com∣mon in Trigonometry, which hath obliged us to make a Table thereof, it serveth us to measure all sorts of Tri∣angles, even Spherical. We have in Opticks some Propositions founded on those

••ouch Lines, as when we are to determine the part of a Globe which is enlightned. The Theorie of the Phases of the Moon is esta∣blished on the same Doctrine, as also that Celebrated Problem by which Hipparchus findeth the distance of the Sun, by the diffe∣rence of the true and apparent Quadra∣tures. In the Gnomonicks, the Italian and Babylonian hours, are often Tangent Lines. We measure the Earth by a Line which toucheth its Surface. And we take in the Art of Navigation, a touch Line to be our Horizon.

PROPOSITION XX. THEOREM.

THe Angle of the Center is double to an Angle of the Circumference, which hath the same Arch for Base.

If the Angle ABC which is in the Center, and the Angle ADC which is in the Circumference, have the same Arch AC for Base; the first shall be double to the second, this Proposition hath three Cases; the first is when the Line ABD passeth through the Center B.

Demonstration. The Angle ABC is exteriour in respect of the Triangle BDC: Thence (by the 32^d of the 1^st.) it shall be equal to the Two Angles D, and C; which being equal (by the 5^th of the 1^st,) because the Sides BC, BD, are equal; the Angle ABC shall be double of each.

The Second case is, when an Angle encloseth the other, and the Lines making the same Angles, not meeting each other, as you see in the second figure; the Angle BID is in the Center, and the Angle BAD is at the Circumference. Draw the Line AIC through the Center.

Demonstration. The Angle BIC is double to the Angle BAC; and CID, double to the Angle CAD, (by the pre∣ceding case:) Therefore the Angle BID shall be double to the Angle BAD.

USE.

THere is given ordinarily a practical way to describe a Horizontal Dial, by a single opening of the Compass, which is grounded in part on this Proposition. Secondly, when we would determine the Apogaeon of the Sun, and the excentricity

of his Circle, by Three observations; we suppose that the Angle at the Center is double to the Angle at the Circumference. Ptolomy makes often use of this Propo∣sition, to determine as well the excentricity of the Sun, as the Moon's epicycle. The first Proposition of the Third Book of Tri∣gonometry is grounded on this.

PROPOSITION XXI. THEOREM.

THe Angles which are in the same Segment of a Circle, or that have the same Arch for Base, are equal.

If the Angles BAC, BDC, are in the same Segment of a Circle, greater than a Semicircle; they shall be equal. Draw the Lines BI, CI.

Demonstration. The Angles A and D are each of them half of the Angle BIC, by the preceding Proposition; therefore they are equal. They have also the same Arch BC for Base.

Secondly, let the Angles A and D be in a Segment BAC, less than a semi-circle; they shall notwithstanding be equal.

Demonstration. The Angles of the Triangle ABE are equal to all the Angles of the Triangle DEC: The Angles ECD, ABE, are equal (by the preceding case;) since they are in the same Segment ABCD, greater than a Semi-circle: the Angles in E are like∣wise equal (by the 15^th. of the 1^st.) therefore the Angles A and D shall be equal, which Angles have also the same Arch BFC, for Base.

USE.

* 1.1 IT is proved in Opticks, that the Line BC shall appear equal, being seen from A and D; since it always ap∣peareth under equal Angles.

We make use of this Proposition to de∣scribe a great Circle, without having its Center; For Example, when we would give a Spherical figure, to Brass Cauldrons to the end we may work thereon; and to pollish Prospective or Telescope Glasses. For having made in Iron an Angle BAC equal to that which the Segment ABC contains; and having put in the Points B, and C, two small pins of Iron, if the Triangle BAC be made to move after such a manner, that the Side AB may always touch the Pin

B; and the Side AC, the Pin C: the Point A shall be always in the Circum∣ference of the Circle ABCD. This way of describing a Circle may also serve to make large Astrolabes.

PROPOSITION XXII. THEOREM.

QƲadrilateral figures described in a Circle, have their opposite Angles equal to Two Right.

Let a Quadrilateral or four sided figure, be described in a Circle; in such manner that all its Angles touch the Circumference of the Circle ABCD: I say that its opposite Angles BAD, BCD, are equal to two Right. Draw the Diagonals AC, BD.

Demonstration. All the Angles of the Triangle BAD are equal to Two Right. In the place of its Angle ABD, put the Angle ACD, which is equal thereto (by the 21^st.) as being in the same Segment ABCD; and in the place of its Angle ADB, put the Angle ACB, which is in the same Segment of the Circle BCDA. So then the Angle BAD, and the Angles

ACD, ACB, that is to say the whole Angle BCD, is equal to Two Right.

USE.

PTolomy maketh use of this Proposition, to make the Tables of Chords or Subtendents. I have also made use thereof in Trigonometry, in the Third Book, to prove that the sides of an Obtuse Angled Triangle, hath the same reason amongst themselves, as the Sines of their opposite Angles.

PROPOSITION XXIII. THEOREM.

TWo like Segments of a Circle described on the same Line, are equal.

I call like Segments of a Circle, those which contain equal Angles; and I say that if they be described on the same Line AB; they shall fall one on the other, and shall not surpass each other in any part. For if they did surpass each other, as doth the Segment ACB, the Segment ADB; they would not be like. And to demonstrate it,

draw the Lines ADC, DB, and BC.

Demonstration. The Angle ADB is exteriour in respect of the Triangle BDC: Thence (by the 21^th. of the 1^st.) it is greater than the Angle ACB, and by consequence, the Segments ADB, ACB, containeth unequal Angles, which I call unlike.

PROPOSITION XXIV. THEOREM.

TWo like Segments of Circles de∣scribed on equal Lines, are equal.

If the Segments of Circles AEB, CFD, are like, and if the Lines AB, CD, are equal, they shall be equal.

Demonstration. Let it be imagined that the Line CD be placed on AB, they shall not surpass each other, seeing they are supposed equal; and then the Segments AEB, CED, shall be de∣scribed on the same Line; and they shall then be equal by the preceding Proposi∣tion.

USE.

* 1.2 CƲrved Lined Figures are often re∣duced to Right Lined by this Propo∣sition. As if one should describe Two like Segments of Circles, AEC, ADB, on the equal sides AB, AC, of the Triangle ABC: It is evident that Transposing the Segment AEC, on ADB; the Tri∣angle ABC is equal to the figure ADBCEA.

PROPOSITION XXV. PROBLEM.

TO compleat a Circle whereof we have but a part.

There is given the Arch ABC, and we would compleat the Circle: There needeth but to find its Center. Draw the Lines AB, BC, and having Divi∣ded them in the middle in D and E; draw their Perpendiculars DI, EI, which shall meet each other in the Point I, the Center of the Circle.

Demonstration. The Center is in the Line DI (by the Coroll. of the 1^st.) It is also in EI; it is then in the Point I.

USE.

* 1.3 THis Proposition cometh very fre∣quently in use; it might be pro∣pounded another way; as to inscribe a Triangle in a Circle; or to make a Circle pass through three given points, provided they be not placed in a streight Line. Let be proposed the Points A, B, C; put the Point of the Compass in C, and at what opening soever describe Two Arks F and E. Transport the foot of the Compass to B; and with the same opening, describe Two Arcks to cut the former in E and F. Describe on B as Center, at what opening soever, the Arches H and G; and at the same opening of the Compass, describe on the Center A Two Arks, to cut the same in G and H. Draw the Lines FE, and GH, which will cut each other in the Point D the Center of the Circle. The Demonstration is evident enough; for if you had drawn the Lines AB, BC, you would have divided them equally, and Per∣pendicularly by so doing. This is very necessary to describe Astrolabes, and to

compleat Circles, of which we have but a part. That Astronomical Proposition which teacheth to find the Apogeum, and the excentricity of the Suns Circle, requires this Proposition. We often make use there of in the Treatise of cutting of stones.

PROPOSITION XXVI. THEOREM.

THe equal Angles which are at the Center, or at the Circumference of equal Circles, have for Base equal Arks.

If the equal Angles D and I are in the Center of equal Circles ABC, EFG; the Arks BC, FG, shall be equal. For if the Ark BC was greater or lesser than the Ark FG: seeing that the Arks are the measure of the Angles; the Angle D would be greater or lesser than the Angle I.

And if the equal Angles A and E be in the Circumference of the equal Circles; the Angles D and I which are the double of the Angles A and E, being also equal; the Arks BC, FG, shall be also equal.

PROPOSITION XXVII. THEOREM.

THe Angles which are either in the Center or in the Circumference of equal Circles, and which hath equal Arks for Base, are also equal.

If the Angles D and I are in the Centers of equal Circles; and if they have for Base equal Arcks BC, FG, they shall be equal, because that their measures BC, FG, are equal; and if the Angles A and E be in the Circum∣ference of equal Circles, have for Base equal Arks BC, EG, the Angles in the Center shall be equal; and they being their halfs (by the 20^th.) shall be also equal.

PROPOSITION XXVIII. THEOREM.

EQual Lines in equal Circles, corre∣spond to equal Arks.

If the Line BC, EF, are applyed in equal Circles, ABC, DEF; they shall be Chords of equal Arks, BC, EF.

Draw the Lines AB, AC, DE, EF.

Demonstration. In the Triangles ABC, DEF, the Sides AB, AC, DE, EF, are equal, being the Semi-Diameters of equal Circles; the Bases BC, EF, are supposed equal: thence (by the 8^th. of the 1^st.) the Angles A and D shall be equal; and (by the 16^th,) the Arks BC, EF, shall be also equal.

PROPOSITION XXIX. THEOREM.

LInes which subtend equal Arcks, in equal Circles, are equal.

If the Lines BC, EF, subtend equal Arks BC, EF, in equal Circles; those Lines are equal.

Demonstration. The Arks BC, EF, are equal, and parts of equal Circles: therefore (by the 27^th,) the Angles A and D shall be equal. So then in the Triangles CAB, EDF, the Sides AB, AC, DE, DF, being equal, as also the Angles A and D; the Bases BC, EF, shall be equal (by the 4^th. of the 1^st.)

USE.

THeodosius demonstrateth by the 28^th, and 29^th, that the Arks of the Circles of the Italian and Babylonian hours comprehended between Two Parallels are equal. We demonstrate also after the same manner, that the Arks of Circles of Astronomical hours comprehended be∣tween Two Parallels to the Equator, are equal; these Propositions come almost con∣tinually in use in spherical Trigonometry, as also in Gnomonicks.

PROPOSITION XXX. PROBLEM.

TO divide an Ark of a Circle into Two equal parts.

It is proposed to Divide the Ark AEB into Two equal parts, put the Foot of the Compass in the Point A, make Two Arks F and G; then trans∣porting the Compass without opening or shutting it, to the Point B: describe two Arks cutting the former in F and G; the Line GF, will cut the Ark

AB equally in the Point E. Draw the Line AB.

Demonstration. You divide the Line AB equally by the construction, for imagine the Lines AF, BF, AG, BG, which I have not drawn, lest I should imbroil the figure; the Triangles FGA, FGB, have all their Sides equal; so then (by the 8^th. of the 1^st.) the Angles AFD, BFD, are equal. Moreover the Triangles DFA, DFB, have the Sides DF, common; the Sides AF, BF, equal, and the Angles DFA, DFB, equal; whence (by the 4^th of the 1^st,) the Bases AD, DB, are equal; and the Angles ADF, BDF, are equal. We have then divided the Line AB equally and perpendicularly in the Point D. So then (by the 1^st,) the Center of the Circle is in the Line EG. Let it be the Point C, and let be drawn the Lines CA, CB; all the Sides of the Triangles ACD, BCD, are equal: Thence the Angles ACD, BCD, are equal (by the 8^th of the 1^st;) and (by the 27^th,) the Arks AE, EB, are equal.

USE.

AS we have often need to divide an Ark in the middle, the practice of this Proposition is very ordinarily in use, it is by this means we divide the Mariners Compass into 32 Rumbs: for having drawn Two Diameters, which cut each other at Right Angles; we divide the Circle in Four, and sub-dividing each quarter in the middle, we have Eight parts, and sub-dividing each part twice, we come to Thirty Two parts. We have also occasion of the same practice, to divide a Semi-circle into 180 degrees; and because for the performing the same Division through∣out; we are obliged to divide an Ark into Three, all the Ancient Geometricians have endeavoured to find a method to divide an Angle or an Ark into Three equal parts, but it is not yet found.

PROPOSITION XXXI. THEOREM.

THe Angle which is in a Semi-circle is Right, that which is comprehended in a greater Segment, is Acute; and that in a lesser Segment, is Obtuse.

If the Angle BAC be in a Semi-circle: I demonstrate that it is Right. Draw the Line DA.

Demonstration. The Angle ADB ex∣teriour in respect of the Triangle DAC is equal (by the 32d. of the 1st.) to the Two Interiours DAC, DCA; and those being equal (by the 5th. of the 1st.) seeing the Sides DA, DC, are equal; it shall be double to the Angle DAC. In like manner the Angle ADC is double to the Angle DAB: therefore the Two Angles ADB, ADC, which are equal to Two Right, are double to the Angle BAC, and by consequence the Angle BAC is a Right Angle.

Secondly, the Angle AEC which is in the Segment AEC, is obtuse; for in the Quadrilateral ABCE, the Opposite Angles E and B, are equal to Two Right (by the 22d.) the Angle B is Acute; therefore the Angle E shall be Obtuse.

Thirdly, the Angle B which is in the Segment ABC, greater than a Semi-circle, is Acute; seeing that in the Triangle ABC, the Angle BAC is a Right Angle.

USE.

* 1.4 THe Workmen have drawn from this Proposition the way of trying if their Squares be exact; for having drawn a Semi-circle BAD, they apply the Point A of their Square BAD, on the Circum∣ference of the Circle, and one of its Sides AB, on the Point B of the Diameter; the other Side AD must touch the other Point D, which is the other end of the Diameter.

Ptolomy makes use of this Proposition to make the Table of Subtendants or Chords, of which he hath occasion in Trigonometry.

* 1.5 We have also a practical way to erect a perpendicular on the end of a Line, which is founded on this Proposition. For Ex∣ample, to erect a Perpendicular from the Point A of the Line AB, I put the Foot of the Compass on the Point C, taken at discretion; and extending the other to A, I describe a Circle which may cut the Line AB in the Point B. I draw the Line BCD. It is evident that the Line DA shall be perpendicular to the Line AB; seeing the Angle BAD is in the Semi-circle.

PROPOSITION XXXII. THEOREM.

THe Line which cutteth the Circle at the Point of touching, maketh with the touch Line the Angles equal to those of the Alternate Segments.

Let the Line BD cut the Circle in the Point B, which is the Point where the Line AB doth touch the same: I say that the Angle CBD which the Line BD Comprehendeth with the touch Line ABC, is equal to the Angle E, which is that of the Alternate Segment BED; and that the Angle ABD, is equal to the Angle F, of the Segment BFD.

In the first place, if the Line passeth through the Center, as doth the Line BE: It would make with the touch Line AB, two Right Angles (by the 17th.) and the Angle of the Semi-circles would be also Right (by the preceding.) So the Proposition would be true.

If the Line passeth not through the Center, as doth the Line BD: Draw the Line BE through the Center, and joyn the Line DE:

Demonstration. The Line BE ma∣keth two Right Angles with the touch Line, and all the Angles of the Triangle BDE are equal to Two Right (by the 32d. of the 1st.) So taking away the Right Angles ABE, and D, which is in a Semi-circle, and taking again away the Angle EBD, which is common to both; the Angle CBD shall be equal to the Angle BED.

Thirdly, the Angle ABD is equal to the Angle F, because in the Quadri∣lateral BFDE, which is inscribed in a Circle, the opposite Angles E, and F, are equal to two Right (by the 22d.) the Angles ABD, DBC, are also equal to Two Right (by the 13th. of the 1st.) and the Angle DBC, and E, are equal, as just now I did demonstrate; therefore the Angles ABD, and BFD, are equal.

USE.

THis Proposition is necessary for that which followeth.

PROPOSITION XXXIII. PROBLEM.

TO describe upon a Line, a Segment of a Circle, which shall contain a given Angle.

It is proposed to describe on the Line AB, a Segment of a Circle to contain the Angle C. Make the Angle BAD equal to the Angle C; and draw to AD the Perpendicular AE. Make also the Angle ABF, equal to the Angle BAF; and lastly describe a Circle on the Point F as Center, at the opening BF, or FA, the Segment BEA con∣taineth an Angle equal to the Angle C.

Demonstration. The Angles BAF, ABF, being equal, the Lines FA, FB, are equal (by the 6th.) and the Circle which is described on the Center F passeth through A and B: Now the Angle DAE being Right, the Line DA toucheth the Circle in the Point A, (by the 16th.) therefore the Angle which the Segment BEA compre∣hendeth as the Angle E, is equal to the Angle DAB; that is to say to the Angle C.

But if the Angle was obtuse; we must take the Acute Angle which is its Com∣plement to 180 degrees.

PROPOSITION XXXIV. PROBLEM.

A Circle being given, to cut there-from a Segment to contain an assigned Angle.

To cut from the Circle BCE, a Segment to contain the Angle A. Draw (by the 17th.) the touch Line BD; and make the Angle DBC equal to the Angle A. It is evident (by the 32d.) that the Segment BEC contains an Angle equal to DBC, and by consequence to the Angle A.

USE.

I Have made use of this Proposition to find Geometrically the excentricity of the Annual Circle of the Sun, and his Apogeon, Three observations being given. It is also made use of in Opticks, Two unequal Lines being proposed, to find a Point where they shall appear equal, or under equal

Angles; making on each, Segments which may contain equal Angles.

PROPOSITION XXXV. THEOREM.

IF Two Lines cut each other in a Circle, the Rectangle comprehended under the parts of the one, is equal to the Rectangle comprehended under the parts of the other.

In the first place, if Two Lines cut each other in the Center of the Circle, they shall be equal and divided equally, so then it is evident, that the Rectangle comprehended under the parts of the one, is equal to the Rectangle compre∣hended under the parts of the other.

Secondly, let one of the Lines pass through the Center F; as AC, and divide the Line BD in two equally in the Point E: I say that the Rectangle comprehended under AE, EC, is equal to the Rectangle comprehended under BE, ED, that is to say to the Square of BE. The Line AC is per∣pendicular to BD (by the Third.)

Demonstration. Seeing that the Line AC is divided equally in F, and un∣equally in F; the Rectangle compre∣hended under AE, EC, with the Square of FE, is equal to the Square of FC, or FB, (by the 5th. of the 2d.) Now the Angle E being Right, the Square of FB is equal to the Squares of BE, EF; therefore the Rectangle comprehended under AE, EC, with the Square of EF, is equal to the Squares of BE, EF; and taking away the Square of EF, there remains that the Square of BE, is equal to the Rectangle under BE, ED.

Thirdly, let the Line AB pass through the Center F, and let it divide the Line CD unequally in the Point E: draw FG perpendicularly to CD; and (by the 3d.) the Lines CG, GD, shall be equal.

Demonstration. Seeing the Line AB is divided equally in F, and unequally in E; the Rectangle comprehended under AE, EB, with the Square of EF, is equal to the Square of BF, or FC, (by the 5th. of the 2d.) In the place of EF, put the Squares of FG, GE, which is equal thereto, (by the 47th. of the 1st.) In like manner the

Line CD being equally divided in G, and unequally in E; the Rectangle CED, with the Square of GE, shall be equal to the Square of GC. Add the Square of GF; the Rectangle of CE, ED, with the Squares of GE, FG, shall be equal to the Squares of GC, GF; that is to say (by the 47th. of the 1st.) to the Square of CF. There∣fore the Rectangle AEB, with the Squares of GE, GF; and the Rect∣angle of CE, ED, with the same Squares are equal; and by consequence taking away the same Squares, the Rectangle AEB is equal to the Rect∣angle CFD.

Fourthly, let the Lines CD, HI, cut each other in the Point E, so that neither of them pass through the Center. I say that the Rectangle CED is equal to the Rectangle HEI. For drawing the Line AFB, the Rectangles CED, HEI, are equal to the Rectangle AEB, (by the preceding case.) therefore they are equal.

USE.

ONe might by this Proposition have a practical way to find the Fourth proportional to Three given Lines, or the Third proportional to Two Lines.

PROPOSITION XXXVI. THEOREM.

IF from a Point taken without a Circle, one draw a touch Line, and another Line to cut the Circle, the Square of the Tangent Line shall be equal to the Rect∣angle comprehended under the whole Secant, and under the exteriour Line.

Let from the Point A, taken without the Circle, be drawn a Line AB, to touch the same in B; another AC, or AH, to cut the Circle. The Square of AB shall be equal to the Rectangle com∣prehended under AC, AO; as also to the Rectangle comprehended under AH, AF. If the Secant pass through the Center, as AC; draw the Line EB.

Demonstration. Seeing the Line OC, is divided in the middle in E, and that thereto is added the Line AO; the

Rectangle comprehended under AO, AC, with the Square of OE, or EB, shall be equal to the Square of AE, (by the 6th. of the 2d.) Now the Line AB toucheth the Circle in the Point B: so (by the 17^th.) the Angle B is Right; and (by the 47th. of the 1st.) the Square of AE is equal to the Squares of AB, EB. Therefore the Rectangle under AC, AO, with the Square of EB, is equal to the Squares of AB, EB: and taking away the Squares of EB; the Rectangle under AC, AO, shall be equal to the Square of AB.

Secondly, let the Secant AH not pass through the Center. Draw the Line AH, the perpendicular EG, which shall divide in the middle in G, the Line FH: draw also the Line EF.

Demonstration. The Line FH being divided equally in the point G; and the Line AF being added thereto; the Rectangle comprehended under AH, AF, with the Square of FG, shall be equal to the Square of AG. Add to both the Square of EG; the Rectangle AH, AF, with the Square of FG, GE, that is to say (by the 47th. of the 1st.) with the Square of FE or EB, shall be equal to the Square of AG,

GE, that is to say (by the 47th. of the 1st.) to the Square of AE. Moreover, the Square of AE, (by the same,) is equal to the Square of EB, AB: therefore the Rectangle compprehended under AH, AF, with the Square of BE, is equal to the Squares of BE, AB, and taking from both the Square of BE, the Rectangle comprehended under AH, AF, shall be equal to the Square of AB.

Coroll. 1. If you draw several Secants AC, AH; the Rectangles AC, AO; and AH, AF, shall be equal amongst themselves, seeing that the one and the other are equal to the Square of AB.

Coroll. 2. If there be drawn Two Tangents AB, AI, they shall be equal because their Squares are equal to the same Rectangle AC, AO; and by consequence amongst themselves; as also the Lines.

PROPOSITION XXXVII. THEOREM.

IF the Rectangle comprehended under the Secant, and under the exteriour Line, is equal to a Line which falls on the Circle; that Line toucheth the same.

Let be drawn the Secant AC or AH, and let the Rectangle AC, AO, or the Rectangle AH, AF, be equal to the Square of the Line AB; that Line shall touch the Circle: draw the touch Line AI (by the 17th.) and the Line IE.

Demonstration. Seeing the Line AI toucheth the Circle, the Rectangle AC, AO; or AH, AF, shall be equal to the Square of AI. Now the Square of AB, is supposed equal to each Rectangle: therefore the Squares of AI, and of AB, are equal; and by consequence the Lines AI, AB. So then the Triangles ABE, AEI, which have all their Sides equal to each other, shall be equi∣angled (by the 8th. of the 1st.) and seeing the Angle AIE is Right (by the 17th.) the Line AI being a touch Line, the Angle ABE shall be Right, and the

Line AB a touch Line also (by the 16th.)

USE.

* 1.6 MAurolicus maketh use of this Propo∣sition to find the Diameter of the Earth. For looking from the top of a Moun∣tain OA, to the extremity of the Earth, along the Line BA; he observeth the Angle OAB, which the Line AB, maketh with a plumb Line AC: and he concludes the Length of the Line AB, by a Trigo∣nometrical Calculation. He Multiplieth AB by AB to have its Square, which he divides by AO, the height of the Moun∣tain: from which having taken away AO there remains OC, the Diameter of the Earth, this Proposition serveth also to prove the Fifth Proposition of the Third Book of Trigonometry.

Notes

• * 1.1

  Prop. XXI.
• * 1.2

  Use 24.
• * 1.3

  Use 25.
• * 1.4

  Use 31.
• * 1.5

  Use 31.
• * 1.6

  Prop. XXXVII.