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And if from each of these equals there be taken the angle DCB, which is common to both, the remaining angle Bad will be equal to the remaining angle BFD; or, which is the same thing, the two angles DFE, EFB will be equal to the two angles DAE, EAB, which is impoffible (I. 16.)

The circumference of the circle, therefore, cannot pass through the point F; and the same may be demonstrated of any other point in the line EA, except the point A; whence a circle may be described about the quadrilateral ABCD, as was to be shewn.

PRO P. XX. THEOREM.

Segments of circles, which stand upon equal chords, and contain equal angles, are equal to each other.

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Let ACB, DFE be two segments of circles, which stand upon the equal chords AB, DE, and contain equal angles ; then will those segments be equal to each other.

For let the segment DFE be applied to the segment ACB, so that the point D may fall upon the point A, and the line DE upon the line AB.

Then, since de is equal to AB (by Hyp.), the point E will fall upon the point B, and the two fegments will coincide with each other. H

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For if they do not, there must be some point, in the circumference of one of them, which will fall either within or without the other.

Let the point F, in the circumference of the circle DFE, be that point, which suppofe to fall at G within the circle ACB; and draw the lines AGC, BC and DG,

Then, since the outward angle AGB, of the triangle BCG, is greater than the inward opposite angle GCB, it will also be greater than the angle DFE, which is equal to GCB, or ACB (by Hyp.).

But the angle AGB is also equal to the angle dfe, her cause the fegments in which they stand are identical; whence they are equal and unequal at the fame time, which is abfurd.

The point F, therefore, cannot fall within the circle ACB; and in the fame manner it may be shewn that it cannot fall without it; consequently the segments must coincide, and be equal to each other.

Q. E. D. COROLL. Segments of circles, which stand upon equal chords, and contain equal angles, have equal circumferences.

PROP. XXI. THEOREM.

In equal circles, equal angles stand upon equal arcs, whether they be at the centres or circumferences; and if the arcs be equal, the angles will be equal.

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Let ABC, der be two equal circles, having the angles AGB, DHE, at their centres, equal to each other, as also the angles ACB, DFE, at their circumferences; then will the arc AKB be equal to the arc DLE.

For, join the points ab, de: then, since the circles are equal to each other (by Hyp.), their radii and circumferences will also be equal (III. 5.)

And, since the two fides AG, GB of the triangle ABG, are equal to the two fides DH, HE of the triangle DEH, and the angle AGB to the angle DHE (by Hyp.), their bases AB, DE will likewise be equal (1.4.)

The chord AB, therefore, being equal to the chord DE, and the angle ACB to the angle die (by Hypo), the arc BCA will also be equal to the arc efd (Cor. III. 20.)

But since the whole circumference of the circle ABC, is equal to the whole circumference of the circle der, and the arc BCA to the arc EFD, the arc AKB will also be equal to the arc DLE.

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Again,

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Again, let the arc AKB be equal to the arc DLE; then will the angle AGB be equal to the angle dhe, and the angle ACB to the angle dfe.

For, if AGB be not equal to DHE, one of them must be greater than the other ; let AGB be the greater ; and make the angle AGK equal to DHE (I. 20.)

Then, fince equal angles stand upon equal arcs (III. 21.), the arc AK will be equal to the arc dle.

But the arc dle is equal to the arc AKB (by Hyp.); whence the arc Ak is also equal to the arc AKB; the less to the greater, which is impoflible.

The angle AGB, therefore, is not greater than the angle DHE ; and in the same manner it may be proved that it cannot be less ; consequently they are equal to each other.

And since angles at the centre are double to those at the circumference, the angle ACB will also be equal to the angle DFE.

Q. E. D.

PRO P. XXII. THEOREM.

In equal circles,

circles, equal chords subtend equal arcs, the greater equal to the greater, and the less to the less; and if the chords be equal the arcs will be equal.

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D

Let ABC, der be two equal circles, in which the chord AB is equal to the chord De; then will the arc

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ACB be equal to the arc dfe, and the arc AKB to the
arc DLE.

For, find G, H, the centres of the circles (III. 1.),
and join GA, GB, HD and HE.

Then, since the circles are equal to each other (hy:
Hyp.) their radii and circumferences will also be equal
(III. 5.)

And, since the fides AG, GB are equal to the fides
DH, He, and the bale AB to the base de (by Hyp.), the
angle AGB will also be equal to the angle DHE (I. 21.)

But equal angles, at the centres of equal circles, stand upon equal arcs (III. 21.); therefore the arc AKB is equal to the arc DLE.

And since the whole circumference of the circle ABC is equal to the whole circumference of the circle der, and the arc AKB to the arc DLE, the arc ACB will also be equal to the arc die,

Again, let ABC, DEF be two equal circles, of which the arc AKB is equal to the arc DLE; then will the chord AB be equal to the chord DE.

For let G, H be the centres of the circles, found as be.
fore ; and join' AG, GB, DH and HE.

Then, since the circles are equal to each other (by
Hyp.), the radii AG, GB will be equal to the radii DH,
HE (III. 5.)

And because the arc AKB is equal to the 'arc DLE (by
Hyp.), the angles AGB, Dhe, at the centres, will be
equal (III. 21.)

Bụt, since the two sides. AG, GB are equal to the two fides DH, He, and the angle AGB to the angle DHE, the base AB will also be equal to the bale DE (1. 4.) Q. E. D.

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H 3

PRO P.

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