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This complete definition of angular an angle at the circumference upon the magnitude is of the greatest importance same base A B: the angle ACB shall in the higher parts of the mathematics, be double of the angle A D B. and may be well illustrated by help of Join D C, and produce it to E. Then, the measuring circumference.

because C A is equal to CD (I. 6.), the With the cen

angle CAD is equal to CDA: theretre C, and radius

fore the angle A CE, which is equal to CA, let there be

CAD, C D A together (I. 19.) is doudescribed a cir

ble of CDA. In like manner it may cle AQ Q, Q3,

be shown that the angle BCE is double and let the dia

of C D B. Therefore the sum or differmeters A Q 2,

ence of the angles ECA, ECB is also Q Q, be drawn

double of the sum or difference of the at right angles

angles C DA, C D B, that is, the angle to one another,

AC B is double of the angle A D B. dividing the whole angular space about the centre into 4 equal angles, each of The adjoined figure which will be measured by a quadrant,* shows that this proof or fourth part of the circumference. is equally applicable

Let us now suppose that the radius when the angle ACB of the circle, being made to revolve is re-entering. about its centre from the original position CA, is brought successively into Therefore, &c. the positions CQ, CQ,, CQ;, and Cor. 1. Any angle at the circumferthence again, continuing its revolution, ence is measured by half the arc upon a second time into the same positions which it stands. CA, CQ, CQ2, CQs, and so on. Cor. 2. (Euc. iii. 26 and 27, secona Then it is evident that the angular space parts of, and Euc. vi. 33 part of). In through which the radius will have re- the same or in equal circles, equal angles volved, will be, in these successive po- at the circumference stand upon equal sitions, one, two, three right angles; arcs, and conversely; and, generally, any upon returning to A four right angles, ngles at the circumference are as the which is the whole angular space

about arcs upon which they stand. (13. and the point C: and thence again, coming II. 17.) a second time to the same positions CQ, CQ,, CQ3, five, six, seven right angles, and so on : which angular spaces

Prop. 15. (Euc. iii. 21.) will be measured respectively by one, Angles in the same segment of a cirtwo, three quadrants, a whole circum- cle are equal to one another. ference; five, six, seven quadrants, and

For they are halves of the same anso on: and any angular spaces interme- gle, víz. the angle at the centre which diate to these will be measured by cor

stands upon their common base; or, responding arcs intermediate, that is, of which is the same thing, they are meamagnitudes between one and two, two sured by the same arc, viz. the halt of and three, three and four, &c. quadrants. their common base.

Cor. 1. (Euc. ii. 31., first part of.) Prop. 14. (Euc. iii. 20.)

The angle which is in a The angle at the centre of a circle is semicircle is a right andouble of the angle at the circumference gle, for it is measured upon the same base, that is, upon the by half the semi-circumsame part of the circumference.

ference, that is, by a
Let A CB be any angle at the centre quadrant.
C of the circle ABD, and let ADB be

Cor. 2. (Euc. iii. 31
second part of.)
The angle, which is in
a segment greater than
a semicircle, is less than
a right angle; and the
angle, which is in a seg-
ment less than a semi-

circle, is greater than a • Proin the Latin word quadrans, a fourth part. right angle; for the one is measured by

D

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an arc which is greater, and the other line be drawn cutting the circle, the by an arc which is less than a quadrant. angles which it makes with the tangent

Cor. 3. If upon the base of a tri. shall be equal to the angles which are angle there be described a segment of a contained in the alternate segments of circle, the vertex of the triangle shall the circle. fall without, or within, or upon the arc Let the straight line A B touch the of the segment, according as the verti- circle C D E in the point C, and from cal angle of the triangle is less than, or C let there be greater than, or equal to the angle in drawn the straight the segment.

line C D cutting For it may easily be shown, (I. 8 the circle: the anCor. 1.) that if the vertex fall within gle DCA shall the arc of the segment, the vertical be equal to the angle must be greater than the angle of angle in the segthe segment, and if without it, less. ment DFC, and

the angle D C B to the angle in the segPROP. 16.

ment DEC. If any chord be drawn in a circle, the tangent A B, and therefore (2. Cor. 2.)

From C draw CE at right angles to the angles contained in the two opposite segments shall be together equal

to two passing through the centre of the circle?

let CE meet the circumference in E: right angles.

take any point F in the arc of the oppoLet ADB be a cir

site segment, and join CE, ED, DF, cle, and let it be di

FC. Then, because CDE is a semivided by the chord AB

circle, the angle C D E is a right angle into the segments AD

(15. Cor. 1.): therefore the remaining B, AEB: the angles

angles of the triangle CDE (I. 19.), ADB, AEB contained

that is, the angles DEC and D CE, are in these segments shall

together equal to a right angle. But the be together equal to

angles D C B and DCE are likewise two right angles For the angle A D B is measured by fore the two latter angles are equal to

together equal to a right angle: therehalf the arc AEB upon which it stands the two former, and the angle D C B is (14. Cor. 1.), and in like manner the equal to DEC, that is, to the angle in angle A E B is measured by half the arc the alternate segment. AD B. Therefore the angles ADB, And because (16.) the angles in the AEB together are measured by half the two segments are together equal to circumference, that is, by two quadrants, two right angles, that is (I. 2.), to the and are consequently equal to two right angles DCB, DC A, the angle DCA angles.

is equal to the angle in the other seg. Therefore, &c.

ment DFC. Cor. 1. (Euc. iii. 22.) If a quadri- Therefore, &c. lateral figure be inscribed in a circle, Cor. The converse is also true: that either pair of its opposite angles shall is, if from the extremity of a chord there be equal to two right angles.

be drawn a straight line, such that the Cor. 2. And conversely, if the oppo. angles which it makes with the chord site angles of a quadrilateral be together are equal to the angles in the alternate equal to two right angles, a circle may segments of the circle, that straight line be described about it. For, if the cir- must touch the circle. cle described through the three points A, D, B (5. Cor.) were to cut the side

Scholium. BE in any other point than E, suppose The theorem which has been just F, the angles A FB, A D B being equal demonstrated, states no more than is to two right angles, would be equal to the contained in Prop. 15., if the tangent be angles A E B, A D B, and therefore the considered as a chord in which the angle A F B to the angle A EB; where

points of section are coincident. For, as one of them, being exterior, must if the point F be supposed to move (I. 8. Cor. 1.) be greater than the other. up to the point C, the chord CF will

tend more and more to coincide in PROP. 17. (Euc. iii, 32.)

position with the tangent C B. But V a straight line touch a circle, and if E F be joined, then, by Prop. 15, if from the point of contact a straight the angle DCF is always equal to the

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BE

A

B

angle DEF. Therefore, when F coin- arcs, according as the point in which cides with C, that is, when the chord they meet is within or without the C F becomes a tangent at C, the angle circle. DCB is equal to the angle D EC. As Let ABC be a cirthis, however, was a case not contem- cle, and let the plated in the demonstration given of chords A B, CD that proposition, the inference could meet one another not have been directly drawn from it. in the point E: the The proposition (2.) that the tangent angle A E C shall is at right angles to the radius is an be measured by instance of the same kind. Others may half the sun or by be seen in the corollaries of the two half the difference following propositions, and in certain of the arcs, AC, properties of tangents which will be BD, according as found in the next section.

the point E is with

in or without the PROP. 18.

circle. Parallel chords intercept equal arcs ;

Through B draw B F parallel to DC,

and let it meet the circumference in F: and conversely.

then (18.) the arc F C is equal to B D, Let ABC be a F.

and therefore the arc A F is equal to the circle, and let the

sum or to the difference of AC, BD, chords A B, CD be

according as the point E is within or parallel to one ano.

without the circle. But, because B F is ther: the arc AC

parallel to DC, the angle A EC is shall be equal to the arc BD.

equal to A B F (I. 15.); and A B F is

measured by half the arc AF,(14.Cor.1.): Join BC. Then, because AB is therefore the angle A E C is measured parallel to CD, the angle A B C (I. 15.) by half the sum or by half the differis equal to the angle B C D: therefore

ence of the arcs A C, BD, according as (14. Cor. 2.) the arc A C is equal to the the point E is within or without the arc B D.

circle. And conversely, if the arc AC be Therefore, &c equal to the arc BD, the angle When the point E is in the circumA B C will (14. Cor. 2.) be equal to the ference, the result of this proposition angle BCD, and therefore (s. 15.) A B coincides with that of 14. Cor. 1. will be parallel to CD.

Cor. By a similar demonstration Therefore, &c.

(18. Cor.) if a chord meet a tangent in Cor. If one of the chords, as AB, be a point which is not the point of contact, surposed to move parallel to itself until the angle contained by them will be the points A and B in which it cuts the measured by half the difference of the circle coincide, as at E, the same and intercepted arcs. its converse will be true: that is, if a The case of a chord meeting a tanchord and tangent be parallel, they gent in the point of contact, has been shall intercept equal arcs; and con- already contemplated in Prop. 17. It versely.

may be considered, however, as included For, because E F is parallel to CD, under the above rule, the measuring arc the angle FEC is equal (I. 15.) to the in this case being the same by this coangle É CD, which stands upon the arc rollary as by Prop. 17. ED: but, because EF is a tangent, (17.) the same F E C is equal to EDC which stands upon the arc E C. There

Section 3.-Rectangles under the fore the arc EC is equal to ED,

Segments of Chords.
(14. Cor. 2.) And the proof of the con-
verse is similarly varied.

PROP. 20. (Euc. iii. 35.)
PROP. 19.

If two chords of a circle cut one

another, the rectangles under their seg. If two chords of a circle meet one ments terminating in the points of another, the angle contained by them section shall be equal, whether they shall be measured by half the sum, or cut one another within or without the by half the difference of the intercepted circle.

B

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B

Let ABC be a

EC is equal to the rectangle under circle, and let the

A E, E B. chords A B, CD

Therefore, &c. cut, or be produced

The same remark may be made here to cut, one another

as at the end of the preceding proposiin the point E: the

tion: viz., that an easy demonstration rectangle under AE,

is likewise afforded by 1. 39. and I. 36. EB shall be equal

Cor. 1. to the rectangle un

Cor. And hence, conversely, if two der CE, ED. A,

straight lines AB, CE cut one another Join AD, BC.

in a point E, and if the points A, B and Then, because the

C, be so taken, that the square of EC angle EAD is equal

be equal to the rectangle under A E, to the angle ECB

EB, the straight line E C shall touch in the same segment (15.), and that the circle which passes through the the angles at E, which are vertical (I. 3.) points A, B, C as in the upper figure, or coincide as in

Prop. 22. (Euc. vi. B.) the lower, are equal to one another, the triangles AED, CEB are equiangular.

If the vertical or exterior-vertical Therefore (II. 31.) AE:ED::EC: angle of a triangle be bisected by a EB, and (II. 38.) the rectangle under

straight line, which cuts the base, or AE, E B is equal to the rectangle under the base produced, the square of that CE, ED.

straight line shall be equal to the Therefore, &c.

difference of the rectangles under the We may remark that an easy demon- two sides, and under the segments of stration of this proposition is likewise the base, or of the base produced.* afforded by I. 39.; for the rectangles in

Let ABC be a triangle, and let the question are each of them equal to the vertical or exterior-vertical angle be difference of the squares of the radius, bisected by the and of the distance of the point E from straight

line the centre of the circle.

A D, which of Cor. And hence, conversely, if two meets the base straight lines A B C D cut one another or

the base in a point E, and if the points A, B and produced in D: C, D be so taken, that the rectangle

The

square of under A E, E B be equal to the rect- AD shall be angle under CE, ED; the points equal to the A, B, C, D shall lie in the circumference

difference of the same circle.

the rectangles

BA, AC, and
PROP. 21. (Euc. iii. 36.)

BD, DC.
If any chord of a circle be produced

Let AEC be the circle which (5.Cor.) to cut a tangent to the same circle, the passes through the points A, B, C, and square of the

ent shall be equal to let AD be produced to meet the cirthe rectangle under the segments of the cumference in E, and join E C. chord.

Then, because the .angles BAD, Let ABC be a circle, and let the EAC are halves, or supplementary to chord A B be pro

the halves of the bisected angle, they duced to meet the

are equal to one another: also the angle tangent CE in E:

ABD is equal to the angle A EC in the square of CE

the same segment (15.): Therefore, the shall be equal to

triangles BAD, E A C being equianguthe rectangle under

lar, (11.31.) BA : AD :: EA: A C, and AE, E B.

(II. 38.) the rectangle under B A, AC is Join CA, C B. Then, because the equal to the rectangle under EA, A D. angle ECB is equal to the angle EAC Again, because the chords BC, E A in the alternate segment of the circle, cut one another in D (20.), the rect(17.) and that the angle at E is com

• This, as is evident from the enunciation, is a mon to the two triangles E CB, EAC, property not of the circle, Int of a triangle, and these two triangles are equiangular. belongs as such to I. $ 6. The required demonstra

Yion has, however, in this and one or two other it. Therefore (11. 31.) AE:EC::EC:

stances rendered an infringement of the classification E B, and (IL. 38. Cor. 1.) the square of unavoidable.

of

E

as

33

angle under BD, DC is equal to the EAC, B A D being equianguar, BA : rectangle under ED, DA: therefore, AD:: EA; AC-(II 31.), and (II. 38.) the difference of the rectangles under the rectangle under BA, AC is equal BA, A C and B D, D C is equal to the to the rectangle under E A, A D. difference of the rectangles under E A, Therefore, &c. AD, and ED, DA, that is, to the square Cor. If two triangles be inscribed in of AD (1. 31.).

the same, or in equal circles, the rect. Therefore, &.c.

angle under the two sides of the one, It should be observed in the case of shall be to the rectangle under the two exterior bisection (see the lower figure), sides of the other, as the perpendicular, that the bisecting line AD must, if pro- which is drawn from the vertex to the duced, cut the circumference in a second base of the one, to the perpendicular point E, in all cases in which it cuts the which is drawn from the vertex to the base B C produced in a point D; that base of the other (II. 35.). is, in all cases in which the sides A B,

PROP. 24, AC are unequal. For when A B is equal to A C, the angles ABC, ACB are

If a quadrilateral be inscribed in a likewise equal" (I. 6.), and therefore circle, its diagonals shall be to one an. (I. 19. and I. ax. 5.) equal to the halves other as the sums of the rectangles under of the exterior angle: therefore, the the sides adjacent to their extremities.

Let ACBD be a angle CAD beingequal to ACB, AD is parallel to BC (1. 15.), and the same quadrilateral figure, CAD being equal to the angle ABC inscribed in the circle in the alternate segment, AD touches A B C and AB, CD the circle in A (17. Cor.). But, when its diagonals: A B one of the sides, as A B, is greater than shall be to CD, a the other, the angle ACB is also greater the sum of CAXA DC than ABC (I. 9.); therefore the angle and C BXB D to the CAD, which (I. 19.) is equal to half the

sum of AC X CB, sum of the two ABC, ACB, is less than

and A D X D B. ACB, and greater than ABC; and

Let A B, and C D cut one another in because the angle CAD is not equal to

the point E: and, first, let A B cut CD ACB, AD is not parallel to BC (1.15.) ;

at right angles. Then, because ACD, and because the same CAD is not equal

BCD, CA B, and D A B are triangles to ABC, that is to the angle in the alter

inscribed in the same circle, the pernate segment, A D does not touch the

pendiculars A E, B E, C E and D E, are circle in A, but cuts it and meets the to one another as the rectangles A CX circumference in a second point E, as

A D, B C B D, CAC B, and D AX was observed.

DB: therefore, (II. 25. Cor. 3.) the sum

of A E and B E, that is A B, is to the PROP, 23. (Euc. vi. C.)

sum of CE and D Е, that is C D, as the If a triangle be inscribed in a circle,

sum of AC XAD, and B CXBD to and if a perpendicular be drawn from the sum of C AXC B and D AND B. the vertex to the buse; the rectangle In the next place, let A B cut CD, but under the two sides shull be equal to the not at right angles : and let the perrectangle under the perpendicular and pendiculars A a, Bb, Cc, and D d be the diameter of the circle.

drawn. Then, as before, it may be Let A B C be a triangle inscribed in shown that A a+B b is to Cc+D d, the circle ABC; from A draw AD

as ACX AD+B C x B D to CAX perpendicular to BC, and

CB+DARD B. But, because the triAE through the centre of

angles A Ea, B Eb, CEC, D Ed are the circle to meet the B

equiangular, A a, Bb, C c, and D d are circumference in E: the

to one another as A E, BE,CE,

and rectangle under BA, AC

DE (II. 31.). Therefore, A a + B b is shall be equal to the rect

to Cc+Dd, as A E+B E to CE+DE, angie under EA, AD.

that is, as AB to CD. Therefore, Join CE. Then, because ACE is a

(II. 12.) A B:CD :: A CXAD+BC semicircle, (15. Cor. 1.) the angle ACE BD:CAXCB+DARD B. is a right angle: but ADB is likewise

Therefore, &c. a right angle, and the angle A EC is

Prop. 25. (Euc. vi. D. equal to the angle A B D in the same If a quadrilateral be inscribed in a segment (15.); therefore, the triangles circle, the rectangle under its diagonals

C

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