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This complete definition of angular magnitude is of the greatest importance in the higher parts of the mathematics, and may be well illustrated by help of the measuring circumference.

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With the centre C, and radius CA, let there be described a circle A Q Q, Q3, and let the diameters A Q2 Q Q, be drawn at right angles to one another, dividing the whole angular space about the centre into 4 equal angles, each of which will be measured by a quadrant,* or fourth part of the circumference.

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Let us now suppose that the radius of the circle, being made to revolve about its centre from the original position CA, is brought successively into the positions C Q, C Q,, C Q, and thence again, continuing its revolution, a second time into the same positions CA, CQ, CQ2, CQ,, and so on. Then it is evident that the angular space through which the radius will have revolved, will be, in these successive positions, one, two, three right angles; upon returning to A four right angles, which is the whole angular space about the point C: and thence again, coming a second time to the same positions C Q, C Q2, C Q., five, six, seven right angles, and so on: which angular spaces will be measured respectively by one, two, three quadrants, a whole circumference; five, six, seven quadrants, and so on: and any angular spaces intermediate to these will be measured by corresponding arcs intermediate, that is, of magnitudes between one and two, two and three, three and four, &c. quadrants.

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PROP. 14. (EUc. iii. 20.)

The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.

Let A CB be any angle at the centre C of the circle A BD, and let AD B be

D

an angle at the circumference upon the same base A B: the angle ACB shall be double of the angle A D B.

Join D C, and produce it to E. Then, because C A is equal to CD (I. 6.), the angle CAD is equal to CD A: therefore the angle ACE, which is equal to CAD, CDA together (I. 19.) is double of CD A. In like manner it may be shown that the angle B C E is double of C D B. Therefore the sum or difference of the angles E C A, ECB is also double of the sum or difference of the angles CD A, C D B, that is, the angle ACB is double of the angle A D B.

• From the Latin word quadrans, a fourth part.

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For they are halves of the same angle, viz. the angle at the centre which stands upon their common base; or, which is the same thing, they are measured by the same arc, viz. the halt of their common base.

Cor. 1. (Euc. iii. 31., first part of.) The angle which is in a semicircle is a right angle, for it is measured by half the semi-circumference, that is, by a quadrant.

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 segment less than a semicircle, is greater than a

right angle; for the one is measured by

an arc which is greater, and the other by an arc which is less than a quadrant. Cor. 3. If upon the base of a triangle there be described a segment of a circle, the vertex of the triangle shall fall without, or within, or upon the arc of the segment, according as the vertical angle of the triangle is less than, or greater than, or equal to the angle in the segment.

For it may easily be shown, (I. 8 Cor. 1.) that if the vertex fall within the arc of the segment, the vertical angle must be greater than the angle of the segment, and if without it, less.

PROP. 16.

If any chord be drawn in a circle, the angles contained in the two opposite segments shall be together equal to two right angles.

Let A D B be a circle, and let it be divided by the chord AB into the segments A D B, AEB: the angles ADB, AEB contained in these segments shall be together equal to two right angles

For the angle ADB is measured by half the arc AEB upon hich it stands (14. Cor. 1.), and in like manner the angle AEB is measured by half the arc AD B. Therefore the angles ADB, AEB together are measured by half the circumference, that is, by two quadrants, and are consequently equal to two right angles.

Therefore, &c.

Cor. 1. (Euc. iii. 22.) If a quadrilateral figure be inscribed in a circle, either pair of its opposite angles shall be equal to two right angles.

Cor. 2. And conversely, if the opposite angles of a quadrilateral be together equal to two right angles, a circle may be described about it. For, if the circle described through the three points A, D, B (5. Cor.) were to cut the side BE in any other point than E, suppose F, the angles AFB, ADB being equal to two right angles, would be equal to the angles A E B, A D B, and therefore the angle AFB to the angle A EB; whereas one of them, being exterior, must (I. 8. Cor. 1.) be greater than the other.

PROP. 17. (Euc. iii. 32.)

If a straight line touch a circle, and if from the point of contact a straight

line be drawn cutting the circle, the angles which it makes with the tangent shall be equal to the angles which are contained in the alternate segments of the circle.

Let the straight line A B touch the circle C D E in the point C, and from C let there be drawn the straight line CD cutting the circle: the angle DCA shall be equal to the angle in the seg ment DFC, and the angle D C B to the angle in the segment D E C.

C

B

From C draw CE at right angles to the tangent A B, and therefore (2. Cor. 2.) passing through the centre of the circle let CE meet the circumference in E: take any point F in the arc of the opposite segment, and join CE, ED, DF, FC. Then, because C D E is a semicircle, the angle CD E is a right angle (15. Cor. 1.): therefore the remaining angles of the triangle CDE (I. 19.), that is, the angles DEC and DCE, are together equal to a right angle. But the angles D C B and D CE are likewise together equal to a right angle: therefore the two latter angles are equal to the two former, and the angle D C B is equal to DEC, that is, to the angle in the alternate segment.

And because (16.) the angles in the two segments are together equal to two right angles, that is (I. 2.), to the angles DCB, DC A, the angle D CA is equal to the angle in the other segment D F C.

Therefore, &c.

Cor. The converse is also true that is, if from the extremity of a chord there be drawn a straight line, such that the angles which it makes with the chord are equal to the angles in the alternate segments of the circle, that straight line must touch the circle.

Scholium.

The theorem which has been just demonstrated, states no more than is contained in Prop. 15., if the tangent be considered as a chord in which the points of section are coincident. For, if the point F be supposed to move up to the point C, the chord CF will

tend more and more to coincide in But position with the tangent C B. if E F be joined, then, by Prop. 15, the angle DCF is always equal to the

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As

angle DEF. Therefore, when F coin-
cides with C, that is, when the chord
C F becomes a tangent at C, the angle
DCB is equal to the angle D E C.
this, however, was a case not contem-
plated in the demonstration given of
that proposition, the inference could
not have been directly drawn from it.
The proposition (2.) that the tangent
is at right angles to the radius is an
instance of the same kind. Others may
be seen in the corollaries of the two
following propositions, and in certain
properties of tangents which will be
found in the next section.

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arcs, according as the point in which they meet is within or without the circle.

B

Join B C. Then, because AB is parallel to CD, the angle ABC (I. 15.) is equal to the angle B CD: therefore (14. Cor. 2.) the arc A C is equal to the arc B D.

And conversely, if the arc AC be equal to the arc BD, the angle A B C will (14. Cor. 2.) be equal to the angle B C D, and therefore (Î. 15.) A B will be parallel to C D.

Therefore, &c.

Let ABC be a circle, and let the chords AB, CD

Cor. If one of the chords, as AB, be supposed to move parallel to itself until the points A and B in which it cuts the circle coincide, as at E, the same and its converse will be true: that is, if a chord and tangent be parallel, they shall intercept equal arcs; and conversely.

For, because EF is parallel to CD, the angle FEC is equal (I. 15.) to the angle E C D, which stands upon the arc ED: but, because EF is a tangent, (17.) the same FEC is equal to EDC which stands upon the arc E C. Therefore the arc EC is equal to ED, (14. Cor. 2.) And the proof of the converse is similarly varied.

meet one another
in the point E: the
angle AEC shall
be measured by
half the sun or by
half the difference
of the arcs, AC,
BD, according as
the point E is with-
in or without the
circle.

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PROP. 19.

If two chords of a circle meet one another, the angle contained by them shall be measured by half the sum, or by half the difference of the intercepted

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Through B draw B F parallel to DC, and let it meet the circumference in F: then (18.) the arc FC is equal to BD, and therefore the arc A F is equal to the sum or to the difference of AC, BD, according as the point E is within or without the circle. But, because B F is parallel to DC, the angle AEC is equal to ABF (I. 15.); and ABF is measured by half the arc AF, (14.Cor.1.): therefore the angle A E C is measured by half the sum or by half the difference of the arcs A C, B D, according as the point E is within or without the circle.

Therefore, &c

When the point E is in the circumference, the result of this proposition coincides with that of 14. Cor. 1.

Cor. By a similar demonstration (18. Cor.) if a chord meet a tangent in a point which is not the point of contact, the angle contained by them will be measured by half the difference of the intercepted arcs.

The case of a chord meeting a tangent in the point of contact, has been already contemplated in Prop. 17. It may be considered, however, as included under the above rule, the measuring arc in this case being the same by this corollary as by Prop. 17.

SECTION 3.-Rectangles under the
Segments of Chords.

PROP. 20. (EUc. iii. 35.)

If two chords of a circle cut one another, the rectangles under their seg. ments terminating in the points of section shall be equal, whether they cut one another within or without the circle.

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Join AD, BC. Then, because the angle EAD is equal to the angle ECB in the same segment (15.), and that the angles at E, which are vertical (I. 3.) as in the upper figure, or coincide as in the lower, are equal to one another, the triangles A E D, CEB are equiangular. Therefore (II. 31.) AE: ED::EC: E B, and (II. 38.) the rectangle under A E, EB is equal to the rectangle under

CE, ED.

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EC is equal to the rectangle under A E, EB.

Therefore, &c.

The same remark may be made here as at the end of the preceding proposition: viz., that an easy demonstration is likewise afforded by I. 39. and I. 36. Cor. 1.

Cor. And hence, conversely, if two straight lines A B, CE cut one another in a point E, and if the points A, B and C, be so taken, that the square of E C be equal to the rectangle under A E, EB, the straight line E C shall touch the circle_which passes through the points A, B, C

PROP. 22. (Euc. vi. B.)

If the vertical or exterior-vertical angle of a triangle be bisected by a straight line, which cuts the base, or the base produced, the square of that straight line shall be equal to the difference of the rectangles under the two sides, and under the segments of the base, or of the base produced.*

Let A B C be a triangle, and let the vertical or exterior-vertical angle be bisected by the straight line A D, which B meets the base

or the base

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D

E

Let A E C be the circle which (5. Cor.) passes through the points A, B, C, and let AD be produced to meet the circumference in E, and join E C.

Then, because the angles BAD, EAC are halves, or supplementary to the halves of the bisected angle, they are equal to one another: also the angle ABD is equal to the angle A EC in the same segment (15.): therefore, the triangles BA D, E A C being equiangular, (II. 31.) BA: AD :: EA: A C, and (II. 38.) the rectangle under B A, AC is equal to the rectangle under E A, A D. Again, because the chords B C, E A cut one another in D (20.), the rect

This, as is evident from the enunciation, is a property not of the circle, bnt of a triangle, and belongs as such to I. § 6. The required demonstra tion has, however, in this and one or two other i stances rendered an infringement of the classification unavoidable.

angle under BD, DC is equal to the rectangle under ED, DA: therefore, the difference of the rectangles under BA, A C and B D, DC is equal to the difference of the rectangles under E A, A D, and ED, DA, that is, to the square of AD (I. 31.).

Therefore, &c.

It should be observed in the case of exterior bisection (see the lower figure), that the bisecting line AD must, if produced, cut the circumference in a second point E, in all cases in which it cuts the base B C produced in a point D; that is, in all cases in which the sides A B, AC are unequal. For when AB is equal to A C, the angles AB C, ACB are likewise equal (I. 6.), and therefore (I. 19. and I. ax. 5.) equal to the halves of the exterior angle: therefore, the angle CAD beingequal to A CB, AD is parallel to BC (I. 15.), and the same CAD being equal to the angle ABC in the alternate segment, AD touches the circle in A (17. Cor.). But, when one of the sides, as A B, is greater than the other, the angle ACB is also greater than ABC (I. 9.); therefore the angle CAD, which (I. 19.) is equal to half the sum of the two ABC, ACB, is less than ACB, and greater than ABC; and because the angle CAD is not equal to ACB, AD is not parallel to BC (I. 15.) ; and because the same CAD is not equal to ABC, that is to the angle in the alternate segment, A D does not touch the circle in A, but cuts it and meets the circumference in a second point E, as was observed.

PROP. 23. (EUC. vi. C.)

If a triangle be inscribed in a circle, and if a perpendicular be drawn from the vertex to the base; the rectangle under the two sides shall be equal to the rectangle under the perpendicular and the diameter of the circle.

Let A B C be a triangle inscribed in the circle ABC; from A draw AD perpendicular to BC, and AE through the centre of the circle to meet the circumference in E: the rectangle under BA, AC shall be equal to the rectangle under EA, AD.

B

Join C E. Then, because ACE is a semicircle, (15. Cor. 1.) the angle ACE is a right angle: but ADB is likewise a right angle, and the angle AEC is equal to the angle A B D in the same segment (15.); therefore, the triangles

EAC, BAD being equiangular, BA : AD::EA: AC-(II. 31.), and (II. 38.) the rectangle under BA, AC is equal to the rectangle under E A, A D. Therefore, &c.

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Cor. If two triangles be inscribed in the same, or in equal circles, the rectangle under the two sides of the one, shall be to the rectangle under the two sides of the other, as the perpendicular, which is drawn from the vertex to the base of the one, to the perpendicular which is drawn from the vertex to the base of the other (II. 35.). PROP. 24.

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Let A B, and CD cut one another in the point E: and, first, let A B cut CD at right angles. Then, because A CD, BCD, CAB, and D A B are triangles inscribed in the same circle, the perpendiculars A E, BE, CE and D E, are to one another as the rectangles A Cx AD, BCX BD, CAXC B, and D Ax DB: therefore, (II. 25. Cor. 3.) the sum of AE and B E, that is A B, is to the sum of C E and D E, that is CD, as the sum of A CxAD, and BCX BD to the sum of CAXCB and D AXD B. In the next place, let A B cut C D, but not at right angles: and let the perpendiculars A a, Bb, Cc, and Dd be drawn. Then, as before, it may be shown that Aa+Bb is to Cc+Dd, as ACX AD+BCX BD to CAX CB+DAXD B. But, because the triangles A Ea, BEb, CEC, DEd are equiangular, A a, Bb, C c, and D d are to one another as A E, BE,CE, and DE (II. 31.). Therefore, Aa + Bb is to Cc+Dd, as AE+BE to CE+DE, that is, as AB to CD. Therefore, (II. 12.) A B: CD::ACxAD+BC xBD: CAXCB+DAXD B. Therefore, &c.

PROP. 25. (Euc. vi. D. circle, the rectangle under its diagonals If a quadrilateral be inscribed in a

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