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the sphere and cylinder; through the centre I, join AI, BI. Let FIH be parallel to AD or BC, EIG and K L parallel to AB or DC, the base of the cylinder; K L meeting BI in M, and the circular section of the sphere in N.

Then, if the plane HFBC be conceived to revolve round the line HF as an axis, the square FG will describe the cylinder AG, the quadrant IFG will describe the hemisphere EFG, and the triangle IFB will describe the cone AIB; and, in the rotation, KL, KN, and K M, will describe corresponding sections of those solids, all of which have the common altitude FI,

Now, as I F is equal to FB, by similar triangles KI is equal to KM, and IN is equal to IG or KL, and (Theorem XXXI., Cor.) IN2=IK2 + K N2; or KL2 = K M2 + K N2; or the square of the longest radius of these circular sections is equal to the sum of the squares of the two others. And as circles are to each other as the squares of their radii (Theorem LXI., Cor.), the circle whose radius is K L is equal to the sum of those whose radii are respectively K M and KN; or the section of the cylinder is equal to the sum of the corresponding sections of the sphere and cone; and hence, as the altitudes are the same, the cylinder is equal to the sphere and cone together. But (Theorem LXXVIII., Cor. 2) the cone is onethird of the cylinder having the same base and altitude; therefore a sphere is two-thirds of the cylinder whose base and altitude respectively are equal to the diameter of the sphere.

Cor. 1. Spheres are to each other as the cubes of their diameters, for they are like parts of their circumscribing cylinders.

PART II.

SPHERICAL GEOMETRY.

DEFINITIONS.

1. The circles of a sphere whose planes pass through the centre are called great circles; and those whose planes do not pass through the centre are called less circles of the sphere.

2. The pole of a circle, is a point on the surface of the sphere equidistant from every point in the circumference of the circle.

3. A spherical angle is an angle formed by the surface of a sphere by the arcs of two great circles which intersect each other, and it is the same as the inclination of the planes of the circles, or as the angle formed by the tangents of the ares at the point of intersection.

4. A spherical triangle is a figure formed on the surface of a sphere by the intersection of three planes which meet in the centre of the sphere.

THEOREM LXXXIII.-Any two great circles as ADC, ABC (fig. 101, p. III.), mutually bi

sect each other.

For the centre of the sphere being in the plane of each circle, is in their line of common section, which line, being a straight line (Theorem LXIII.), is therefore a diameter; and hence A D C, and ABC, are semicircles.

THEOREM LXXXIV.-The distance of a great circle, A B C from its pole, D (fig. 101, p. III.), is a quadrant.

For ADC is a semicircle (Theorem LXXXIII.), and (Def. 2, SPHERICAL GEOMETRY) D is equidistant from every part of A BC; A is therefore equidistant from the points A and C, and consequently AD and DC are quadrants.

THEOREM LXXXV.-If two great circles, as BA, CA (fig. 102, p. III.), intersect each other in A, on the surface of a sphere whose centre is D; and if BC be an arc of a great circle whose pole is A, then BC is the measure of the spheri-. cal angle ABC.

For, join AD, BD, and C D, then as A is the pole of BC, A B and A Care quadrants, and the angles ADB and ADC are right angles, and consequently (PLANES Def. 3) the plane angle B DC is the angle made by the planes ADB and ADC, or A is equal to the spherical angle ABC (SPHERICAL GEOMETRY Def. 3). But the angle BDC is measured by the arc BC, hence the equal angle BAC is measured by the same are BC.

Cor. 1. As AD is perpendicular to BD) and DC, it is perpendicular to the plane BDC (Theorem LXIV.), therefore the planes ADB and ADC, which pass through AD, are also perpendicular to the plane BDC. Hence the spherical angles A B C and ACB are right angles.

Cor. 2. Great circles whose planes are at right angles to the plane of another great circle, meet in the poles of that circle.

Cor. 3. Great circles which are at right angles to each other, pass each through the poles of the other; and if one circle pass through the pole of another, it cuts that great circle at right angles.

THEOREM LXXXVI.-Any two sides, as AB and BC, of a triangle as A BC (fig. 103, p. III.) are together greater than the third side AC.

For if I be the centre of the sphere, then the solid angle D is contained by three plane angles ADB, BDC, and ACD, any two of which are greater than the third (Theorem LXXIII). But these angles are measured respectively by the arcs A B, BC, and AC, hence any two of these arcs as AB and BD are together greater than the third arc A C.

THEOREM LXXXVII.-The three sides of a spherical triangle, as A BC (fig. 103, p. III.), are together less than a circle.

For the plane angles which form the solid angle D are altogether less than four right angles (Theorem LXXIV.), therefore the arcs AB, BC, and AC, which measure those plane angles are altogether less than a circle.

THEOREM LXXXVIII.-In isosceles spherical triangles, the angles opposite the equal sides are equal; and if two angles of a spherical triangle are equal, the sides which are opposite those angles are equal.

Let ABC (fig. 104, p. III.) be a spherical triangle, having two sides A B and BC equal, and let D be the centre of the sphere. Let BE and EC he tangents to AB and AC, and BF, CF two tangents drawn from B and C in the plane DBC, and intersecting each other in F, and join FE. Then the angles DBE and DC E being right angles (Theorem XLIV.), and the angles E D B and EDC measured by the equal arcs AB and AC being also equal, and the adjacent sides DB and DC, radii of the same

sphere being equal, the triangles EDB and E DC are identical (Theorem II.) and have EB and BC equal; and BF being equal to FC (Theorem LVI. Cor.), and EF common to the two triangles EF Band E FC, those triangles are identical (Theorem V.) and have the angles EBF and ECF (which are equal to the spherical angles ABC and ACD) equal to each other.

Again, the same construction being made, suppose the spherical angles A B C and AC B, or the plane angles E B F and ECF, to be equal; then as DB is at right angles to B E and BF, A is at right angles to the plane BEF (Theorem LXIV); and therefore the plane D BC, which passes through D B, is at right angles to the plane BEF. (Theorem LXXI.). For a like reason the plane DBC is at right angles to the plane ECF, and consequently (Theorem LXXII.) F E, the common section of the planes EBF, ECF, is at right angles to the plane DBC. Hence EFB and E FC are right angles, and therefore as the angles E BF and ECF are equal, and the side BF equal to the side FC, the triangles EBF and ECF are identical (Theorem II). Whence BE is equal to E C, and as DB is equal to D C, and the angles DBC and DCE are equal, the angle ED B is equal to the EDC (Theorem I.); or the arc A B is equal to the arc A C.

THEOREM LXXXIX. In any spherical triangle as ABC (fig. 105, p. III.) the greater side is opposite the greater angle, and the greater angle opposite the greater side.

If A BC be greater than A, let A B D, a part of ABC, be equal to A; then (Theorem LXXXVIII.) A D is equal to D B, and consequently B D and DC are equal to A D and DC, but BD and DC are greater than BC (Theorom LXXXVI.) therefore A D and D C (or A C) are greater than B C.

Again if A C is greater than BC, then if the angle B is not greater than A, it must be equal to it, or less. If it were equal then AC and BC would be equal (Theorem LXXXVIII.), and if it were less then by the preceding part of the proposition AC would be less than BC; both of which conclusions are inconsistent with the given condition that AC is greater than BC. Hence the angle B must be greater than the angle A.

THEOREM XC.-If A, B, C, the angles of the spherical triangle A B C (fig. 106, p. III.), be the poles of three great circles DE, EF, and FD, then D, E, and F, the points where these circles intersect will be the poles of A C, AB, and BC, respectively; and the sides DE, EF, and FD, will be respectively the supplements of the measures of the angles A, B, and C; and the sides AB, BC, and AC, will also respectively be the supplements of the measures of the angles E, F, and D.

For let A B, AC, and BC, be produced both ways, till they meet DE, EF, and D F, in G, H, I, K, L, and M. Then G H is the measure of the angle BAC, KI the measure of ABC, and ML the measure of AC B (Theorem LXXXV). And as A is the pole of D E, the angle AHD is a right angle (Theorem LXXXV. Cor. 1), and for a like reason the angle CLD is a right angle;

VOL. X.

hence (Theorem LXXXV. Cor. 2) D is the pole of LH. In the same way it may be shown that E is the pole of G K, and F the pole of MI; therefore L H is the measure of D, G K the measure of E, and M I the measure of F.

Now as D is the pole of LH, DH is a quadrant, and as E is the pole of GK, EG is a quadrant (Theorem LXXXIV.); therefore E G and DH, or ED and GH, together are a semicircle. Hence DE is the supplement of GH, the measure of the angle BAC. In a similar way it may be shown that EF is the supplement of the measure ABC, and that DF is the supplement of the measure of AC B.

Again, because A is the pole of DE, AG is a quadrant, and, because B is the pole of E, F is a quadrant. Hence AG and BK, or AB and KG, together are equal to a semicircle; or A B is the supplement of G K, the measure of the angle E. And in the same way it may be shown that AC is the supplement of the measure of D, and BC the supplement of the measure of F.

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THEOREM XCI.-The sum of the interior angles of a spherical triangle is greater than two, and less than six right angles.

For (fig. 106, p. III.) the measures of the angles A, B, C, together with the sides of the supplemental triangle DEF are equal to three semicircles; and as (Theorem LXXXVII.) the three sides of the triangle DEF, are less than four semicircles, the measures of the angles A, B, and C, are greater than two semicircles; or those angles are greater than two right angles.

And, as the interior and exterior angles of any triangle are equal to six right angles, the interior angles alone must be less than six right angles.

THEOREM XCII.-If AGFB (fig. 107, p. III.) be the circumference of a great circle, of which D is the pole, and C any other point on the surface of the sphere, the greatest arc of a circle that can be drawn from C to the circle AGFB is that passing through D, and those which are nearer ADC are greater than those which are more remote.

For, as CA passes through the pole of AF B, the planes ACB and AF B are at right angles; therefore C E drawn in the plane AC B, perpendicular to AB the common section of the two planes, is perpendicular to A E, GE, and FE, which it meets in the plane AFB; and (Theorem XXXVI.) AE is greater than GE, and GE greater than FE; and the right-angled triangles AEC, GEC, FEC, having EC common, the hypothenuse AC is greater than GC, and GC greater than FC. Hence the arc AC is greater than the arc GC, and the arc GC greater than the arc FC.

THEOREM XCIII.—In any right-angled spherical triangle, the sides containing the right angle are of the same affection as their opposite angles; that is, if the sides are greater or less than quadrants, the opposite angles are greater or less than right angles; and, conversely, if the angles are greater or less than right angles, the opposite sides are greater or less than quadrants.

Let ABC (fig. 108, p. III.) be a spherical triangle, right-angled at A; produce A ̊C, A B, till

K

they meet in G, and bisect the semicircles ABG and ACG in E and F. Then F will be the pole of ABG and E the pole of AFG; join CE, then CE will be a quadrant (Theorem LXXXIV.) and ECA will be a right angle (Theorem LXXXV. Cor. 3.). Hence when AB is less than A E, or less than a quadrant, the opposite angle ACB is less than a right angle.

Again, let ADC be a spherical triangle, right angled at A, having AD greater than a quadrant, then the angle DCA is greater than the angle ECA, or greater than a right angle.

The converse may be demonstrated in a simi

lar manner.

THEOREM XCIV.-In the right-angled spherical triangle, if the sides which contain the right angle are of the same affection, the hypothenuse, or the side opposite the right angle, is acute; but if they be of different affections, the hypothenuse is obtuse.

For when AC and AB (fig. 108, p. III.) are each less than a quadrant, CB being farther from CFG than CE, is less than CE (Theorem XCII.), and therefore less than a quadrant. But if AD be greater, and AC less than a quadrant, then CD, being nearer CFG than CE, is greater than CE, or greater than a quadrant.

Again, in the right-angled triangle CGB, where CG and GB are each greater than a quadrant, CB being farther from CFG than CE is less than a quadrant.

Cor. 1. Hence, conversely, if the hypothenuse of a right-angled spherical triangle is greater than a quadrant, the sides about the right angle are of different affections; and, if the hypothe nuse is less than a quadrant, the sides are of the same affection.

THEOREM XCV.-In any spherical triangle, as ABC (fig. 109, p. III.), if the perpendicular BD fall within the triangle, the angles A and C are of the same affection; but, if the perpendicular fall without the triangle, the angles at the base are of different affections.

For when BD falls within the triangle, the angles A and C of the right-angled triangles ADB and BDC, being each of the same affection with BD (Theorem XCIII.), are of the same affection with each other.

But, when the perpendicular falls without the triangle, the angles DAB and DC B, being each of the same affection with BD, are of the same affection with each other; hence BAC and BCA are of different affections.

PART III.
PRACTICAL GEOMETRY.

PROBLEM I. To bisect a straight line, as AB (fig. 110. p. III.).

From the points A and B, as centres, with any radius greater than half A B, describe arcs cutting each other in n and m ; join n m, and AB will be bisected in C, the point in which A is cut by mn. For join An, Am, Bn, and Bm; then because An, and A m, are respectively equal to Bn and Bm, and n m is common to both triangles Anm and Bnm, the angles Anm and B nm are equal. And hence, as An and Bn are equal, C is common to both triangles, and the angle

An C equal to the angle Bn C, the remaining sides AC and BC are equal (Theorem I.), or AB is bisected in C.

PROBLEM II. To bisect a given angle, as ABC (fig. 111, p. III.).

From the angular point B with any radius describe the arc AC, and from the points A and C, with the same or any other radius, describe arcs intersecting in n. Join B n, and A will bisect the angle ABC.

For An and Cn being joined, then (Theorem V.) the triangles ABn and n BC are equal in every respect, and therefore the corresponding angles AB n and CB n are equal, or the angle ABC intersected by the line B n.

PROBLEM III. From a given point C in a given line as AB (figs. 112 and 113, p. III.) to draw a perpendicular.

1. When the point C is near the middle of the line (fig. 112, p. III.) On each side of C take any two equal distances Cn, Cm; and from n and m, with any radius greater than Cn or Cm, describe arcs cutting each other in s. Join s C, then that line will be the required perpendicular.

For (Theorem V.) the angles AC s, and BC s, are equal, and therefore (Def. 7.) C s is perpendicular to AB.

2. When the point C. is near the end of the line (fig. 113, p. III.). Take any point, o, as a centre, and with the radius o C describe an are cutting AB in m and C. Through m and o, draw the line m o n, cutting the arc in n. Join n C, and it will be the perpendicular required.

For the angle m Cn, being in a semicircle, is a right angle (Prop. 37, Cor. 3,) therefore n C is perpendicular to A B.

Or from any scale of equal parts take Cm equal to four parts, and with C as a centre and radius equal to two or three parts, describe an arc; and with m as a centre, and radius equal to five of the same equal parts, describe an arc, cutting the preceding one in n, join n C, which will be perpendicular to A B.

For 524232; hence (Theorem LIV Cor. 2), AC n is a right angle.

PROBLEM IV. From a given point C, out of a given line A B, to draw a line perpendicular to A B (figs. 114 & 115, p. III.)

1. When the given point is nearly opposite the middle of the line, as in fig. 114. Take any point o, on the other side of A B, and from the centre C, with the radius C o, describe an arc, cutting A B in m and n; bisect n m in G, and join CG; then CG is perpendicular to AB. For Cn and Cm are equal, and Gn and Gm are equal, and GC common to the triangles CnG and Cm G, therefore the angles CG n and C Gm are equal (Theorem V.), or CG is perpendicular to A B.

2. When the point is nearly opposite the end of the line, as in fig. 115. Draw any line Cm, from C to AB; bisect Cm in n, and with centre n and radius Cn or mn describe a circle, cutting AB in G; then, if CG be joined, the line will be perpendicular to A B. For the angle CG m being in a semicircle is a right angle.

PROBLEM V.-At a given point D, in a given line DE, to make an angle equal to a given angle A B C (fig. 116, p. III.)

From B as a centre with any radius, describe the arc nm, cutting BA, BC, in the points m, n; and from D as a centre, with the same radius, describe the arc rs; take the distance mn, and apply it to the arc rs, from r to s. Draw DF through D and s, then the angle EDF will be equal to the angle ABC, as is evident from Theorem V.

PROBLEM VI.-Through a given point C (fig. 117, p. III.) to draw a line parallel to A B.

From C to AB draw any line CD; then through C draw CE, making the angle ECD equal to the alternate angle CDA, then (Theorem XVI.) C E is parallel to A B.

PROBLEM VII.-To draw a line, as FC (fig. 118, p. III.), parallel to another line A B, and at the distance of CD from A.

At any point E in AB, draw E F perpendicular to A B, and equal to CD, and through F draw F G parallel to A B, and A will, as is evident, be the required line.

PROBLEM VIII. To divide a line as AB(fig. 119, p. III.) into n equal parts.

Through one extremity A, draw any line AC, and on it take n + 1 equal parts, D being the termination of the n+1th part, I that of the nth part, and F that of the n-1th part. Join DB, and produce it till BE is equal to BD, and join EF, cutting A B in P, then BP is the nth part of A B. For FI being equal to I B, and EB to B D, FE and IB are parallel, therefore A B BP :: AL IF: 1 n; therefore BP is the nth part of A B.

PROBLEM IX.-To find the centre of a circle ADBC (fig. 120, p. III.)

Draw any chord A B, and A, bisect it with the perpendicular CD; bisect C D in O, and O will be the centre of the circle.

For if the centre is in C D, it must be in O, the middle of it. If it is not in CD let it be at F, and join F E, F A, and F B; then A F will be equal to F B, A E to E B, and EF common to the two triangles A FE and BFE, which (Theorem V.) are therefore identical, and consequently the angles A EF and BEF, being equal, will be right angles, and each equal therefore to the angle B E C, which is impossible. Hence the centre is in C D, and consequently it is in the point O.

PROBLEM X.-To describe the circumference of a circle through three given points, A, B, C, (fig. 121, p. III.)

Join AC, BC, and bisect these lines with the perpendiculars DO and E O, and from the point O, with the distance OA, OB, or O C, describe the circle A BC, and it will be the circle required.

For as AD and DO are respectively equal to CD and DO, and the included angles ADO, and C DO, are equal, A O and CO are equal; and for a like reason BO is equal to CO.

Hence O is the centre of a circle passing through A, B, and C.

Note. By this problem a circle may be described about a triangle.

PROBLEM XI.-To inscribe a circle in a triangle, as ABC (fig. 122, p. III.) Bisect any two of the angles as A and B, and the bisecting lines AD and BD will meet in the centre of the circle.

For from D draw DE, D F, and DG, perpendicular to the sides. Then as the angles DAG and DAE are equal, and the right angles DEA and DGA are equal, the angles ADE and ADG are equal; and consequently, as A D is common to both triangles, GD and DE are equal.

In the same way it may be shown that D E and DF are equal; and as the angles E, F, and G, are right angles, a circle described from D as a centre, with any of these equal lines as a radius, will touch the sides in E, F, and G, and it will consequently be inscribed in the triangle.

PROBLEM XII. To make a triangle equal to any rectilineal figure, as ABCDE (fig. 123, P. III.).

Draw A D, and parallel to it draw EF, meeting AB produced in F, then the triangles ADE and AFD are equal (Theorem, XXIV. Cor.) Proceed similarly with DB and CG, and the triangle FDG will obviously be equal to the figures ABCDE.

In the same way the figure may be reduced to a triangle whatever be the number of sides. PROBLEM XIII.-To make a rectangle equal to a given triangle, A B C (fig. 125, p. III.)

Bisect A B in D with the perpendicular D E meeting C F, drawn parallel to AB in E; and draw B F parallel to D E. Then the rectangle DF will be equal to the triangle A B C (Theorem XXV. Cor. 3).

PROBLEM XIV.-To find BD, the side of a square whose area is equal to a rectangle contained by AB and BC (fig. 125, p. III.).

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Upon A C, the sum of A B and B C, describe semicircle, and draw the perpendicular BD, which will be the side of the square. For the rectangle of A B and BC is equal to the rectangle of B D, and the other segment of the chord, of which B D is a part (Theorem LVI.); but (Theorem XLII.) that segment is equal to BD; therefore the rectangle AB BC is equal to the square of BD.

PROBLEM XV.-On a straight line A B (fig. 126, p. III.) to describe the segment of a circle, to contain an angle equal to a given angle.

Make BAC equal to the given angle, bisect AB with the perpendicular DE, meeting A E, a perpendicular to AC in E, then with E as a centre, and E A or E B as a radius, describe the circle AF BG; then (Theorem XL.) the angle BAC is equal to any angle in the segment AGB.

GEOPON'ICAL, adj. GEOPONICS, n. s.

ture.

Fea and Tovog. The doctrine of agricul

just punishment for the enormities of his life. The occasion of his death, however, as narrated by ecclesiastical writers, will not add any stain

Such expressions are frequent in authors geoponical, to his memory. There was in Alexandria a

or such as have treated de re rusticâ.

Browne's Vulgar Errours. GEORGE, n. s. Lat. Georgius. A proper name; a figure of St. George worn by the Knights of the Garter; a name formerly given to a brown loaf, and at present applied to the rolls used at colleges in Oxford.

Look on my george, I am a gentleman; Rate me at what thou wilt.

Shakspeare. Henry VI. Cubbed in a cabin, on a mattrass laid, On a brown george, with lowsy swobbers fed. Dryden. GEORGE I., king of Great Britain. See GREAT BRITAIN and HANOVER.

GEORGE II. See GREAT BRITAIN and HAN

OVER.

GEORGE III. See GREAT BRITAIN. GEORGE (St.), a saint or hero, after whom several orders, both military and religious, are denominated. On some medals of the emperors John and Manuel Comneni, we have the figure of St. George armed, holding a sword or javelin in one hand, and in the other a buckler, with this inscription; an O, and therein

P

a little A, and гE-гIOC, making OʻATIO2O

TEOPTION, O holy George. He is generally represented on horseback; and is highly venerated throughout Armenia, Muscovy, and all the countries which adhere to the Greek church; from the Greek, his worship has long been received into the Latin church; England and Portugal have both chosen him for their patron saint. Great difficulties have been raised about this saint or hero. His very existence has been called in question. Dr. Heylin supposed him only a symbolical device; and Dr. Pettingal asserted him to be a mere Basilidian symbol of victory. The following is Mr. Gibbon's account of this saint. He asserts him to have been an Arian bishop born at Epiphania in Cilicia, in a fuller's shop. From this obscure and servile origin, says this author, he raised himself by the talents of a parasite: and the patrons, whom he assiduously flattered, procured for their worthless dependent a lucrative commission, or contract, to supply the army with bacon. His employment was mean: he rendered it infamous. He accumulated wealth by the basest arts of fraud and corruption; but his malversations were so notorious, that George was compelled to escape from the pursuits of justice. After this disgrace, in which he appears to have saved his fortune at the expence of his honor, he embraced with real or affected zeal the profession of Arianism. From the love, or the ostentation, of learning, he collected a valuable library of history, rhetoric, philosophy, and theology; and the choice of the prevailing faction promoted George of Cappadocia to the throne of Athanasius. His conduct in this station is represented by our historian as polluted by cruelty and avarice, and his death is by him considered as a

place in which the priests used to offer human sacrifices. This place Constantius gave to the church of Alexandria, and George the bishop ordered it to be cleared, to build a Christian church on it. In doing this, they discovered a subterraneous cavern, in which the heathen mysteries had been performed, and in it were many human skulls. These, and other things which they found in the place, the Christians brought out and exposed to public ridicule. The heathens, provoked at this exhibition, took arms, and rushing upon the Christians killed many of them; they also seized the bishop in the church, and put him in prison. The next day they despatched him; and then, fastening his body to a camel, dragged it about the streets all day, and in the evening they burnt it and the camel together. This fate, according to Sozomen, the bishop owed in part to his haughtiness while he was in favor with Constantius; but he ascribes it chiefly to the inveteracy of the heathens, whose superstitions George had been very active in abolishing. But Mr. Gibbon gives a different turn to the affair of George's murder, and relates it with different circumstances. The Pagans,' says he, 'excited his devout avarice; and the rich temples of Alexandria were either pillaged or insulted by the haughty prelate, who exclaimed, in a loud and threatening tone, 'How long will these sepulchres be permitted to stand?" Under the reign of Constantius, he was expelled by the fury, or rather by the justice of the people; and it was not without a violent struggle that the civil and military powers of the state could restore his authority and gratify his revenge. The messenger who proclaimed at Alexandria the accession of Julian announced the downfal of the archbishop. George, with two of his obsequious ministers, count Diodorus and Dracontius master of the mint, was ignominiously dragged in chains to the public prison. At the end of twenty-four days the prison was forced open by the rage of a superstitious multitude, impatient of the tedious forms of judicial proceedings. The enemies of gods and men expired under their cruel insults; the lifeless bodies of the archbishop and his associates were carried in triumph through the streets on the back of a camel; and the inactivity of the Athanasian party was esteemed a shining example of evangelical patience. The remains of these guilty wretches were thrown into the sea; and the popular leaders of the tumult declared their resolution to disappoint the devotion of the Christians, and to intercept the future honors of these martyrs who had been punished, like their predecessors, by the enemies of their religion. The fears of the Pagans were just and their precautions ineffectual. The meritorious death of the archbishop obliterated the memory of his life. The rival of Athanasius was dear and sacred to the Arians, and the seeming conversion of these sectaries introduced his worship into the bosom of the Catholic church. The odious stranger, disguising every circumstance of time and place,

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