all the calamities that befall a large city, a great fire is perhaps the most appalling to behold. Notwithstanding the trained army of the fire brigade, always ready and willing to encounter danger wherever a fire occurs, yet frequently the most disastrous, yes, often the most melancholy results are attendant on a great fire. Human life is often sacrificed, every attempt to resist the devouring element being fruitless. Once a fire gets ahead it is mostly irresistible, from the nature of the materials of a building, or the combustible nature of the contents of a shop or warehouse. The wind may be blowing strong and the flames carried to adjacent edifices; the water may not be in sufficient quantity near to the burning building; some or all of the firemen may be engaged at another fire in a distant part of the town; the fire may break out and remain unperceived until the efforts to restrain its fury arrive too late. These and many other contingencies, unprovided for because unforeseen, may tend to give the fire the mastery over all its skilled opponents.

Of the results attendant on a great fire the following may be briefly enumerated. In the first place, there is the destruction of property which it may have taken years, nay generations, to amass; and though the insurance may to a certain extent recoup the capitalist, there is no such provision for the men employed by the firm whose factory or store is burned down. These in general have made "no provision for the rainy day," and being suddenly and unexpectedly unemployed, much suffering and destitution generally follow. The clerks, the time-keepers, the carters, the shopmen, the travellers, and even the night-watchman, all suffer greatly from the burning of a factory. The burning of a theatre is perhaps still more calamitous, the number of persons employed being often greater, and less likely to get immediate engagements elsewhere. As a rule those employed in theatres only allow life to weigh gently on them. Their habits and profession accustom them to look upon the bright side of things, their minds are of a liberal type, their pockets as light

as their spirits. Without having any practical knowledge of the fact, we venture to say the employés at theatres have few accounts in the savings banks. They are all high-minded and generous, always ready to give their gratuitous services to help a brother in need.

In modern times the most disastrous fire was the burning down of a large portion of the city of Chicago, which rendered homeless some thirty thousand people. With the usual recuperative powers of Americans, the streets were soon rebuilt in greater splendour than the former ones. Next came the calamitous fire of Quebec, by which it was roughly calculated at the time twelve thousand people became homeless. The great metropolis London has had, until quite recently, only ordinary fires to chronicle. The burning down of the Alhambra Theatre on Thursday, the 7th of last month, with the destruction of the block of warehouses in Wood-street, Cheapside, on the following day, are calamities with most deplorable results. As to the origin of either fire nothing whatever is known. The former was discovered at one, the latter at three o'clock in the morning. The theatre being a detached building, the flames were pretty easily restrained from proceeding beyond it. However, it required a'l the well-directed efforts of the fire brigade to prevent the Wood-street fires extending beyond the block of buildings first attacked, where it is computed damage to the amount of two millions sterling has been done, but fortunately without any loss of life. In no place in the world, perhaps, could such a rich and varied assortment of goods packed within so small an area have been destroyed. Had a strong wind been blowing, we might have to record a second great fire of London like that of 1666.

The month of December, 1882, will long be remembered for the vast number of disastrous fires. In addition to that at Hampton Court Palace and that in Cheapside, we have had a match factory burnt down in Belfast, with a lamentable loss of life, consisting of four persons.

To Publishers.-We have from day to day letters asking what text-books we recommend on various subjects, particularly for matriculation in the Universities. Only a few publishers have as yet sent in their books. It will be our object always to recommend the best books" without fear or favour." ED.

Falmouth. We have received dozens of essays as well as yours, which for the present are held over. Two classes of essays we prefer (1) anticipatory, as that on "A Great Fire." (2) those on subjects lately given. Writing 64.

GEOMETRY.

1. On a given right line, construct a rectangle which shall be equal in area to a given rectangle.

Let AB be the given right line, and AD the given rectangle, and suppose one side AE of the rectangle AD to be in the same straight line as AB. Through B draw MK parallel to C, and produce DC till it meets MK in K. Through KA draw KH, meeting DE, produced in H. Through H draw HM parallel to AB, meeting CA produced in L, and KM in M. Then ABML is a rectangle, equal in area to the rectangle AD.

By construction, the lines DH, CL, and KM are paralel, as also the lines DK, EB, and HM, and the line HK is a diameter of the parallelogram DKMH; therefore the complements AM and AD of the parallelograms CB and EL are equal. (I. 43.)

Since the angle CAE is a right angle, each of the angles BAL and ABM is a right angle (I. 15, 29), and consequently the parallelogram AM is a rectangle.

Thus the rectangle AM has been described on the right line AB, and it is equal to the g'ven rectangle AD. (Q.E.F.)

2. Frove that the line joining the middle points of any two sides of a triangle is parallel to the third side; and also that the triangle formed by the lines joining the middle points of the three sides is one-fourth of the given triangle.

Let ABC be the triangle, and DEH the middle roints of the sides. Join DE, DH, EH, BE, and DC.

Since AE-EC, the triangle ADE=the triangle CDE.

Since AD DB, the triangle ADE=the triangle BED.

Therefore the triangle CDE the triangle BED. The triangles CDE and BED are therefore between the same parallels (I. 39), and DE is parallel to BC. (Q.E.D.)

Since DB is one-half of AB, the triangle DBE is one-half of the triangle ABE; and since AE is one-half of AC, the triangle ABE is onehalf of the triangle ABC.

The triangle DBE is consequently one-fourth of the triangle ABC, therefore the triangle DHE is also one-fourth of ABC. (Q.E.D.)

3. In any triangle, prove that the square on the side subtending an acute angle is less than the sum of the squares on the sides containing that angle by twice the rectangle contained by either of these sides, and the intercept between the acute angle and the foot of the perpendicular let fall from the opposite angle. This is Euclid, II. 18.

4. Prove that the sum of the squares on two sides of a triangle is double the sum of the squares on half the base and on the line joining the vertex to the middle point of the base.

Let ABC be the triangle and D the middle point of BC. Join AD, and from A draw AP perpendicular to BC.

5. Prove that the difference of the square on any two lines is equal to the rectangle contained by the sum of the lines and their difference.

Let AB, AC, be the two lines, then ACAB (AB+AC) (AC-AB). On AB construct the square ABLK, and on AC construct the square ACHD. Produce KL to M. The difference of the squares DC and KB is equal to the sum of the rectangles DM and MB.

But DH AC, and HM-HC-MC-HCKA-AC-AB. Therefore the rectangle DM = AC.BC.

Since MC AB, the rectangle MB=AB.BC. Therefore the sum of the rectangles DM and MB is equal to AC.BC+AB.BC, or equal to (AC+ AB). BC. (II. 1.)

Bat BC-AC-AB, therefore the difference of the squares on AC and AB is equal to (AC+ AB).BC, or equal to (AC+AB) (AC-AB). (Q.E.D.)

6. If two circles intersect, prove that the right line joining their centres produced if necessary, bisects the right line joining the points of section of the circles, and is at right angles to it.

Let ECD and HCD be two circles intersecting at C and D, and have their centres at A and B.

In the triangles CAB and DAB, the three sides AC, CB, BA of the one are equal to the three sides AD, DB, BA, of the other respectively; therefore the angle CAB is equal to the angle DAB. (1.8.)

In the triangle ACK and ADK, the side AK is common to the triangles, the side AC of the oneis equal to the side AD of the other, and the angle CAK has been proved equal to the angle DAK; therefore the side CK is equal to the side DK, and the angle CKA is equal to the angle DKA. (I. 4.)

Since CK DK, CD is bisected in K; and since the angle CKA equals the angle DKA; each of these angles is a right angle (Def. 10), and CD is at right angles to AB. (Q.E.D.)

7. On a given right line describe a segment of a circle containing an angle equal to a given. angle. (Euclid, III. 32.)

8. Hence, or otherwise, find a point O within a triangle ABC such that the angles AOB, AOC, and BOC shall be equal. Point out when this construction becomes impossible.

On AC describe a segment of a circle containing an angle of 120 degrees. On AB describe a similar segment. Then the point where the. segments intersect is the point required.

For the point O being by hypothesis within the triangle, the three angles AOB, AOC, and BOC must together equal 360 degrees, and consequently each of them must equal 120 degrees. By the construction the angles 40C and AOB each equal 120 degrees, and conseqrently the remaining angle BOC must also. equal 120 degrees.

When AB lies wholly within the segment AOC, the construction becomes impossible, for the segments do not then intersect within the triangle.

9. Inscribe a circle in a given triangle. This is Euclid IV. 4.

Show also how to describe a circle touching one of the sides and the other two sides produced. Let ABC be the triangle, it is required to describe a circle touching BC, and AB, and AC produced.

lines BD, CD, mee'ing each other in the point D,

AE, BC, and AG in the points E, F, and G respectively.

In the triangles EBD and FBD, the side BD is common to the triangles, the angles EBD and FBD are equal by construction, and the angles BED and BFD are both right angles and consequently equal; therefore, the side DE is equal to the side DF. (I. 26.)

In the same manner it may be shown that DG is equal to DF.

DG, DF, and DE are, therefore, equal to one another, and the circle described from the centre D, with the radius DF, must pass through E and G.

The circle must also touch the lines AE, BC, and AG, in the points E, F, and G respectively, because the radii DE, DF, DG, are by construction respectively at right angles to the lines AE, BC, and AG. (III. 16).

The circle EFG consequently touches the side BC, and the sides AB and AC produced.

(Q. E. F.) 10. On a given right line describe an equilateral and equiangular pentagon.

Let CD be the given straight line. Describe an isosceles triangle FG II, having each of the angles G, H, double the angle F. On CD describe a segment of a circle CBAED, which

The Royal Military Academy, Woolwich. The following in order of merit, and with the total number of marks obtained, are declared by the Civil Service Commissioners to be the successful candidates at the Examination held in November and December, 1882: S. H. Powell, 8,604 marks; L. P. Chapman, 8,521; W. C. Hedley, 8,181; 0. H. Stoehr, 7,837; E A. Edgell, 7,705; W. K. Hardy, 7,350; H. A. A. Livingstone, 6,977; H. O. Blackall, 6,941; H. J. Croftor, 6,764; A. W. Medley, 6,736; W. A. Liddell, 6,717; A. Robinson, 6,663; W. Ewbank, 6,491; T. M. Dickinson, 6,160; A. M. A. Harris, 6,060; A. L. Schreiber, 6,036; R. G. Merriman, 5,946; G. H. Harrison, 5,868; C. R. Stevens, 5,852; F. H. E. Brouncker, 5,817; F. de B.

Draw CB, making the angle BCA equal to the angle ICE. Join CB, BA, AE, ED. Then CBAED is the pentagon required.

By construction the angles BCA, ACE, and ECD are equal, and are also each equal to the angle contained by the segment CBAED; therefore the arcs BA, AE, ED, and DC are equal. (III. 26.) Since each of the above angles is equal to one-fifth of two right angles, each of these four arcs is one-fifth of the whole circumference of the circle CBAED. (III. 20.) Consequently the remaining are BC is equal to one-fifth of the circumference. Therefore the five arcs BA, AE, ED, DC, and CB are equal; therefore, the five corresponding chords are equal (III. 29); therefore, the pentagon is equilateral.

Each of the angles, as AED, stands upon an arc, as ABCD, which is three-fifths of the whole circumference; the angles are, therefore, equal to OLe another (III. 27), and the pentagon is equiangular.

The pentagon ABCDE, described on CD, is consequently both equilateral and equiangalar. (Q. E. F.)

Young, 5,655; H. B. Williams, 5,598; C. G. Watson, 5,570; E. P. Lambert, 5,561; R. E. L. Radcliffe, 5,512; H. W. lles, 5,446; H. J. Sherwood, 5,444; W. Lamport, 5,297; P. T. Cooper, 5,260; G. W. H. Jago, 5,179; C. Lyon, 5,156; E. H. Bland, 5,113; F. N. Friend, 5,088; H. F. Askwith, 5,044; G. T. F. Walker, 4 995; J. A. S. Tulloch, 4,952; A. E. Harrisor, 4,945; E. R. B. Stokes-Roberts, 4,926; G. F. White, 4,920; E. E. Norris, 4,875; T. B. Wood, 4,826; M. H. B. Raby, 4,820; P. Foster, 4,794; H. I. Wilson, 4,778; M. Peake, 4,760; St. J. L. H. Du P. Taylor, 4,743; T. B. Moore, 4,685; F. E. Freeth, 4,632; H. M. Birley, 4,594; E. C. Cameron, 4,576. Some of the papers will appear in our next issue.