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Exercise 2.-To draw a straight line through a given point that shall be parallel to a given straight line; by means of the preceding corollary.

PROP. XXVIII. THEOREM.

In equal circles, equal straight lines cut off equal arcs, the greater equal to the greater, and the less to the less.

Let ABC and DEF be equal circles, and BC and EF equal straight lines in them, which cut off the two greater arcs BAC and EDF, and the two less arcs BGC and EHF. The greater arc B A C is equal to the greater are EDF, and the less arc B G C to the less arc EHF.

Take K and L, the centres of the circles (III. 1), and join B K, KC, EL, and LF.

A

K

D

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E

H

Because the circles A B C and DEF are equal, the straight lines drawn from their centres (III. Def. 1) are equal. Therefore BK and KC are equal to EL and LF, each to each. But the base BC is equal (Hyp.) to the base EF. B Therefore the angle BKC is equal (I. 8) to the angle ELF. Because equal angles stand upon equal arcs (III. 26) the arc BGC is equal to the are EHF. But the whole circumference ABC is equal (Hyp.) to the whole circumference EDF. Therefore the remaining arc BAC, is equal (I. Ax. 3) to the remaining arc EDF. Therefore, in equal circles, &c. Q. E. D.

G

In this and the following propositions, it is assumed by Euclid, that the straight lines or chords are less than the diameter. Dr. Thomson, in his edition, adapts the enunciation and demonstration of both to the case of the diameter also; but this seems to be an unnecessary refinement, as the diameter evidently bisects the circles, making the arcs equal, and conversely.

Corollary.-In the same circle, equal chords cut off equal arcs, the greater equal to the greater, and the less to the less.

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In equal circles, equal arcs are subtended by equal straight lines.

Let A B C and D E F be equal circles, and let the equal arcs B G C and EHF be subtended by the straight lines BC and EF. The straight line BC is equal to the straight line EF. Take K and L (III. 1) the centres of the circles, and join BK, KC, EL, and LF.

Because the arc BGC is equal to the arc EHF, the angle BKC is equal (III. 27) to the angle ELF. Because the circles A B C and DEF are equal, the straight lines from

A

D

K

B

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G

H

their centres are equal (III. Def. 1). Therefore BK and KC are equal to EL and LF, each to each. But the angle BK C is equal to the angle

ELF. Therefore the base BC is equal (I. 4) to the base E F. Therefore, in equal circles, &c. Q. E. D.

Corollary.-In the same circle, equal arcs are subtended by equal straight lines.

PROP. XXX. PROBLEM.

To bisect a given arc; that is, to divide it into two equal parts. Let ADB be the given arc. It is required to bisect it.

Join AB, and bisect (I. 10) it at C. From the point C, draw CD at right angles (I. 11) to AB. The arc ADB is bisected at the point D.

D

C

B

Join AD and DB. Because AC is equal to CB, and C D common to the triangles ACD and BCD, the two sides AC and CD are equal to the two sides BC and CD, each to each. But the angle ACD is equal to the angle BCD, because each of them is a right angle. Therefore the base AD is equal (I. 4) to the base BD. But equal straight lines A cut off equal (III. 28) arcs, the greater equal to the greater, and the less to the less; and AD and DB are each of them less than a semicircle, because DC (III. 1. Cor.) passes through the centre. Therefore the arc AD is equal to the arc DB. Therefore the given arc is bisected at D. Q. E. F.

Corollary. The perpendicular which bisects any chord of a circle, bisects also the arc which it cuts off.

The trisection of any arc of a circle on elementary principles, is as impossible as the trisection of an angle.

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In a circle, the angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Let ABCD be a circle, of which the diameter is B C, and the centre E; and, let any straight line CA divide the circle into the segments ABC and ADC. The angle in the semicircle BAC is a right angle; the angle in the segment ABC, which is greater than a semicircle is less than a right angle; and the angle in the segment AD C, which is less than a semicircle, is greater than a right angle.

E

F

Take any point D in the arc ADC, and join AB, AD, DC, and AC. Also join A E, and produce B'A to F. Because BE is equal (I. Def. 15) to EA, the angle EAB is equal (1. 5) to the angle EBA. Because AE is equal to EC, the angle EAC is equal to the angle B ECA. Therefore the whole angle BAC is equal (I. Ax. 2) to the two angles ABC and AC B. FA C, the exterior angle of the triangle A B C, is equal (I. 32) to the two angles ABC and ACB. Therefore the angle BAC is equal (I. Ax. 1) to the angle FAC, and each of them (I. Def. 10) is a right angle. Therefore the angle BAC in a semicircle is a right angle. Because the two angles ABC and BAC of the triangle ABC are

But

together less than two right angles (I. 17), and BAC has been proved to be a right angle. Therefore A B C is less than a right angle. Wherefore the angle in a segment ABC greater than a semicircle, is less than a right angle. Because ABCD is a quadrilateral figure inscribed in a circle, any two of its opposite angles are equal (III. 22) to two right angles. Therefore the angies A B C and ADC, are equal to two right angles. But ABC has been proved to be less than a right angle. Therefore the angle ADC is greater than a right angle. Wherefore the angle in a segment ADC less than a semicircle is greater than a right angle. Therefore, in a circle, &c. Q. E. D.

COR. From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle: because the angle adjacent to it is equal (I. 32) to the same two; and when the adjacent angles are equal (I. Def. 10) they are right angles.

Dr. Thomson, in his edition, simplifies the demonstration of this proposition, by the application of Prop. XX., after the following manner:-The angle BAC at the circumference is half the two angles at the centre formed by producing A E. But these angles are equal to two right angles. Therefore BAC is a right angle; and it is the angle in a semicircle BA C. The angle BAD is greater than the angle BAC (I. Ax. 9). Therefore the angle BAD is greater than a right angle; and it is the angle in a segment BAD, less than a semicircle. In like manner, by drawing a straight line from the point A, within the angle B A C, it may be shown that the angle in a segment greater than a semicircle is less than a right angle.

Exercise. To draw a perpendicular to a straight line from one of its extremities, without producing it, by means of the first part of this proposition.

PROP. XXXII. THEOREM.

If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle; the angles which this straight line makes with the tangent are equal to the angles in the alternate segments of the circle.

Let the straight line EF touch the circle ABCD at the point B; and from the point B, let the straight line BD be drawn, cutting the circle, and dividing it into the two segments DCB and DAB, of which DCB is less, and DAB greater than a semicircle. The angles which BD makes with the tangent EF are equal to the angles in the alternate segments of the circle; that is, the angle DBF is equal to the angle in the segment DA B, and the angle DBE to the angle in the segment DCB. From the point B, draw BA at right angles (I. 11)

to EF, take any point C in the arc DCB, and join AD, DC, and CB.

B

Because the straight line EF touches the circumference of the circle ABCD at the point B, and BA is drawn at right angles to the tangent from the point of contact B, the centre of the circle is (III. 19) in BA. Therefore the angle ADB in a semicircle (III. 31) E is a right angle. Because the other two angles BAD and ABD, in the triangle AD B, are equal to (I. 32) a right angle, and ABF is (Const.) a right angle. Therefore the angle ABF is equal to the two angles BAD and ABD (I. Ax. 1). From these equals take

away the common angle ABD. Therefore the remaining angle DBF is equal to the remaining angle BAD (I. Ax. 3), which is in the alternate segment of the circle. Because ABCD is a quadrilateral figure in a circle, the opposite angles BAD and BCD are equal (III. 22) to two right angles. But the two angles DBF and DBE are equal (I. 13) to two right angles. Therefore the two angles DBF and DBE are equal (I. Ax. 1) to the two angles BAD and BCD. But the angle DBF has been proved equal to the angle BAD. Therefore the remaining angle DBE is equal (I. Ax. 2) to the remaining angle B CD, which is in the alternate segment of the circle. Wherefore, if a straight line, &c.

Q. E. D.

In Dr. Thomson's edition, the case, in which the straight line cutting the circle passes through the centre, is considered; this is so evident, that it appears to be an unnecessary addition.

Corollary 1.-If the angles which any straight line meeting a circle makes with a chord, be equal to the angles in the alternate segments, that straight line touches the circle.

Corollary 2.-Tangents to a circle at the extremities of the same chord are equal, and make equal angles with it on the same side.

Corollary 3.-The chord which joins the points of contact of parallel tangents is a diameter.

PROP. XXXIII. PROBLEM.

Upon a given straight line to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle.

Let AB be the given straight line, and C the given rectilineal angle. It is required to describe upon the given straight line AB, a segment of a circle, which shall contain an angle equal to the

angle C.

First, let the angle C be a right angle.

Bisect AB in F (1. 10), and from the centre F, at the distance FB, describe the semicircle AHB.

Because

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A

F

B

the segment AHB is a semicircle, the angle AHB is
an angle (III. 31), and therefore it is equal to the given angle C.
Next, let the angle C be an oblique angle.

At the point A, in the straight line AB, make the angle BAD equal

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to the angle C (I. 23); and from the point A, draw A E at right angles (I. 11) to AD. Bisect AB in F (I. 10); and from F, draw FG at right angles (I. 11) to AB. Join GB.

Because AF is equal to FB, and FG common to the triangles AFG and BFG, the two sides AF and FG are equal to the two sides BF and FG, each to each. But the angle AFG is equal (I. Def. 10) to the

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Prom a fan cock to cut of a termed, rich tak contin on cage erudi to a girea restlined. capit.

Let ABC be the given enle and the given rectilined angle. As resized to cut off from the circle ABC a segment that shall contain an ange opal to the given angle D.

Day the night Be EF toushing the dine ABC in the point B L 17), and at the pit B, in the straight line BF, make the ange PBC equal (125) to the angle D. The wigment BAC comzins an angle equal to the given angle b.

Because the straight line EF touches the Groe ABC, and BC is drawn from the point of

ontact B, the angle FBC is equai (III. 32) to the angle in the alternate segment BAC of the circle. But the angle FBC is equal (Const.) to the angle D. Therefore the angle in the segment BAC is equal (I. Arx. 1) to the angle D. Wherefore, from the given circle A B C, the segment AC is cut off, containing an angle equal to the given angle D. QE. P.

Beercise-Through a given point to draw a straight line that shall cut off, from a given circle, a segment containing a given angle.

PROP. XXXV. THEOREM.

If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.

In the circle ABCD let the two straight lines AC and BD, cut one another in the point E. The rectangle contained by AE and EC is equal to the rectangle contained by B E and E D.

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