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LESSONS IN GEOMETRY.-No. XXX.
LECTURES ON EUCLID.
(Continued from page 44.)
BOOK I.

PROPOSITION XXXII. THEOREM.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles. Let A B C be a triangle, fig. 32, No. 1, and let one of its sides BC be produced to n. Then the exterior angle A CD is equal to the two interior and opposite Fig. 32. No. 1. angles CA B and ABC. And the three A interior angles ABC, BCA, and CAB are equal to two right angles. Through the point c draw the straight line, (I. 31) CE, parallel to the side BA.

B

C

E

Because CE is parallel to BA, and c meets them, the angle ACE is equal (I. 29) to the ternate angle BAC. Again, because UE is parallel to AB d BD falls upon them, the exterior angle ECD is equal .29) to the interior and opposite angle ABC. But the angle

ACE was shown to be equal to the angle B AC. Therefore the whole exterior angle ACD is equal (Ax. 2) to the two interior and opposite angles CA B and A BC. To each of these equals, add the angle ACB; the two angles A CD and ACB are therefore equal (Ax. 2) to the three angles CAB, ABC and ACB. But the two angles ACD and ACB are equal (I. 13) to two right angles. Therefore also the three angles CAB, A B C and ACB are equal (4x. 1) to two right angles. Wherefore, if a side of any triangle be produced, etc. Q. E. D.

Corollary 1.- All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Let ABCDE, fig. 32, No. 2, be any rectilineal figure. All the interior angles ABC, BCD, etc., together with four right angles, are equal to twice as many right Fig. 32, No. 2. angles as the figure has sides.

Divide the rectilineal figure ABCDE into as many triangles as the figure has sides, by drawing straight lines from a point F, within the figure to each of its angles.

Because the three interior angles of a triangle are equal (I. 32) to two right angles, and there are as many triangles in the figure as it has sides, all the angles

A

D

EXERCISE II. TO PROPOSITION XXXII.

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equal parts.

of these triangles are equal to twice as many right angles as the figure has sides. But all the angles of these triangles are equal to the interior angles of the figure, viz. A B C, BC D. etc., To trisect a given finite straight line; that is, to divide it into three together with the angles at the point F, which are equal (I 15. Cor. 2) to four right angles. Therefore all the angles of these triangles are equal (4x. 1) to the interior angles of the figure together with four right angles. But it has been proved that all the angles of these triangles are equal to twice as many right angles as the figure has sides. Therefore all the angles of the figure together with four right angles are equal to twice as many right angles as the figure has sides.

Corollary 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. Let fig. 32, No. 3, represent any rectilineal figure.

D

Fig. 32. No. 3.

B

Because the interior angle A B C, and its adjacent exterior angle ABD are (I. 13) together equal to two right angles. Therefore all the interior angles together with all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides. But it has been proved by the foregoing corollary, that all the interior angles together with four right angles are equal to twice as many right angles as the figure has sides. Therefore all the interior angles together with all the exterior angles are equal (4r. 1) to all the interior angles and four right angles. Take from these equals all the interior angles; therefore all the exterior angles of the figure are equal (Ar. 3) to four right angles.

Corollary 1.-If two angles of a triangle be given, the third is given; for it is the difference between their sum and two right angles.

Corollary 2.-If two angles of one triangle be equal to two angies of another triangle, the third angle of the one is equal to the third angle of the other.

Corollary 3.-Every angle of an equilateral triangle is equal to one-third of two right angles, or two-thirds of a right angle. Hence, a right angle can be trisected.

Corollary 4.-If one angle of a triangle be a right angle, the sum of the other two is a right angle. Corollary 5.-If one angle of a triangle be equal to the sum of the other two, it is a right angle.

Corollary 6-If one angle of a triangle be greater than the sum of the other two, it is obtuse; and if less, acute. Corollary 7.-In every isosceles right-angled triangle, each of the acute angles is equal to half a right angle. Corollary 8.-All the interior angles of every quadrilateral figure are together equal to four right angles. This is only a particular case of Euclid's Cor. 1: but it is very necessary to be remembered.

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Because the angle ACD is equal (Hyp.) to the two angles CBA and BAC, add to each of these equals the angle BOA. Then the two angles ACD and BCA, are equal to the three angles BCA, CAB and ABC.

B

Fig.f.
A

D

Let AB, fig. g, be the given straight line. It is required to trisect it; that is, to divide it into three equal parts.

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Fig g

F

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Through the point A draw the straight line A c, making any angle with a B. Produce A c indefinitely to p. From CD cut off c E and EF each, of them equal to A c. Join FB. Through the points c and E. draw CG and EH parallel to F B and meeting A B in G and H (Prop 31). Then A B is divided into three equal parts in the points G and H. Through the points c and E, draw CI and EK, parallel (Prop. 31) to A B, and meeting FB in the points 1 and K; and let C1 meet EH in L. Join G L. Because A G, CL and EK are parallel (Const.), the angles FEK, ECL, and CAG are equal (Prop. 29). And because CG, EL, and FK are parallel (Const.), and AB falls upon them, the angles ACG, CEL and EFK are equal (Prop. 29). And the sides a C, CE and EF of the triangles ACG, ECL and EFK are equal (Const.); therefore, (Prop. 26) the sides AG, CL and EK are equal. But because CL is parallel to GH, and G L meets them, the angles CLG and LG H are equal (Prop. 29); and because co is parallel to LH, and GL meets them, the angles CG L and HLG are equal (Prop. 29); and GL is common to the two triangles GCL and G HL; therefore c L is equal (Prop. 26) to a H. But CL was shown to be equal to AG. Therefore G H is equal (A. 1) to AG. In like manner, by joining L K and H 1, it may be shown that E K is equal to L1, and that LI is equal to H B. But E K was shown to be equal to a o. Therefore H B is equal to A G. But GH is also equal to AG. Therefore, the three straight lines AG, GH, and HB are equal to one another. Wherefore the straight line A B has been trisected in the points G and H. Q. E. F.

Note. By this problem a straight line may be divided into any number of equal parts.

EXERCISE III. TO PROPOSITION XXXII. † Any angle of a triangle is right, acute, or obtuse, according as the straight line drawn from its vertex bisecting the opposite side is equal to, greater than, or less than half that side.

In fig. h, let A B be a triangle, and let a D be drawn from A, bisecting BC in D.

No. 1.

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A A A

CA

D

C

A

This exercise was solved by Q. PRINGLE (Glasgow), as above, and he

But these three angles (Prop. 32) are equal to two right angles.
Therefore the angles ACB and ACD are also equal to two right has the merit of generalising it, so as to make it applicable to the division

angles. Therefore the straight line CD (Prop. 14) is in the same straight line with B C. Q. E. D.

This exercise was solved by Q. PRINGLE (Glasgow), who added other two modes of solution, equally correct; it was also solved by D. H. (Driffield); T. Bocock (Great Warley); E. J. BUEMNER (Carlisle): C. L. HAD FIELD (Bolton-ie Moors); J. H. EASTWOOD (Middleton); and others.

of a straight line into any number of equal parts; a problem which Dr. Thomson postpones till he comes to Prop. xxxiv. It was also solved by WARIN (East Dereham); E. J. BREMNER (Carlisle); J. II. EASTWOOD (Middleton); and others.

This exercise was solved as above by Q. PRINGLE (Glasgow); J. Bocock (Great Warley); C. L. HADFIELD (Bolton-le Moors); H. J. PUGH (Longsight); E. J. BREMNER (Carlisle); D. H. (Driffield); J. H. EASTWOOD (Middleton); and others.

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Because i la greater than be, IVANO DA DAGalam Pron. 18) than they got k Vanka touch, tha woga SARA leas than the wogle WA. Therefore the mye BAC 2 then the two angina a me and won. Infore the angle as (Chu, 6) is moute,

Bed. Let AD be less than bo No. 3. Then the angle BAC la obtuse

Because pA in less than 6, the angle bacin greater than the angle pe A (Prop. 18). For a like reason, the ange bab la greater than the angle DBA. Therefore the angle RACIS Kreater than the angles ane and AGB. Therefore the angle BAD (65%) 6) is obtuse, Wherefore, any angle of a triangle is right, acute, or obtuse, etc, Q. E. D.

EXERCISE IV. TO PROPOSITION XXXII.* The straight line drawn from the vertex of any angle of a triangle Bisecting the apposite side, is equal to, greater than, or less than Auff that side, according as the angle is right, acute, or obtuse, Ing. A, let a n o be a triangle, and let ▲ D be drawn from a bispoting wo in D.

tat. Let the angle A e No. 1 be a right angle. Then A D is equal to pc. For, if not, then AD is either greater or less than we; and, according an AD is greater or less than D C, so is the angle na e acute or obtuse, by the preceding exercise; but it is neither acute nor obtuse by hypothesis; therefore Ap is neither greater nor less than Do-that is, AD is equal

72 in 27 are aga to the fur ga. L. SIDE. these ellas and the ange at and the three angles F ZAP GAP 2 are equate ir uge ¤ the points 7. G. I and Bitte tre mesUL BAFind A FE are eq to two right inges Prog. 10 therefore dr. I the five and at the pointa 7, 6, 7. I 131, are equal to two right ang Wherefore, if the sides of an equilateral ect. QE. D. EXERCISE VL TO PROPOSITION IUL

dud. Let the angle na e No. 2 be acute. Then A D is greater than ne. For, if not, then A D is either equal to, or less than, pc, and, according as A D is equal to, or less than, as the angle ne right or obtuse, by the preceding exercise, but it is neither right nor obtuse by hypothesis; bercire a p is neither equal to, nor less than, Do-that is, AP is greater than pe

det the angle na e No. 3 be obtuse. Then, it may do shown, as in the preceding cases, that Ac is less than a D. Iberviève the straight line drawn, etc. Q. E. D.

FARRCISE V. TO PROPOSITION XXXII.†

If the wider of an equilateral and equiangular kerapy (or six-s
figer i producent fill they weet, the angles formed at the po
of meeting are bojative equal to four right angle.
2, BD, DC, CR and 7, both ways till they meet in the po
In fig. k, let ABCORP be a regular hexagon. Produce the s
6, H, K, L, M and x. Then all the angles at these points are ed
to four right angles.

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ANSWERS TO CORRESPONDENTS.

en apu lateral and equiangular pentagon (or fivebuses" de gradovi till they movt, the angles formed at the many are wyecker equal tu two right angles. SENAMANcos de a regular pentagon. Produce the the English words am, as, and them.

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pronounced as in fate. Final is in the dative and ablative plural! R. 8. T. (Leeds): Pronounce Latin in general like English. Lon nounced as the English word ice; but many adopt the short sound, Aiss. Sim and sis are pronounced like him, hiss; final am, as, and ex The apparent discrepancy be Walker and Smith arises from the fact that the former denotes accen latter quantity. Walker should be followed in pronunciation.

D. D. N. (Slough): Pronounce o, h, to, as hee, he, toe, my as seen wv as tone. It is unnecessary to learn the English-Greek exercises. exercises are well done-particularly the English-Greek. Instead of king he is a general," you should have said "The king himself," etc. H. T. S. H. (Oxford); We fear that his scheme for an air pump, ho ingenious, would not answer; we strongly doubt the applicatility of pistons.-OMEGA: We must try and insert the biography he wants P. E.-J. MC BAIN (Fetter-lane): The Cotton Account is not debit the amount of cash received, but for the amount due by the mercha the cotton purchased on credit or otherwise; and the same acco credited by the amount due for the cotton sold. The reason why the an due are entered, less the discounts, is because the latter are small an tomary, and not worth opening a separate account for; besides, the ess the discounts, are the actual sums for which the cotton was pur or sold, and are, after all, the proper sums to be entered in the 1 The 18o, dental expenses ar sing from the sending of the cott in to ad are, in de manner, added to the cost of the article, as constituting t anunt due for it, or the actual sums for which the cotton was pur As to Ply Cash, its name pibes that it has nothing to Charang dan di Krpenas attending the purchase or sue of Or Ay Sems it moues land wat in trade transactions, it KON 24dae de soi's Labursements which occur in the eval „g-bou pier was made 13 VINCO (Sobo 9206218018 20. De Õen Zadary. Prodarima are very good, but they have c *** ani de pace-me, that we could not insert their

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2nd. That this co-efficient is independent of the pressure, that is, of the original density of the gas. These laws are founded on the supposition that all gases expand equally under an equal increase of temperature, and that the expansion is in direct proportion to the temperature. The method adopted by Gay-Lussac in the determination of these laws was the following: the expansion of air and other gases was measured by an air-thermometer formed of a spherical bulb A, and a capillary stem A B, fig. 183.

The part of the stem between B and c was divided into parts of equal capacity; and the number of these parts which the bulb A contained, was ascertained by weighing the apparatus when full of mercury at 0° Centigrade, and then heating it gently in order to force out of it a little of the mercury. By weighing it again, the weight of the mercury forced out was determined. By cooling the remainder down to 0 Centigrade, a vacuum was produced which showed the volume of the corresponding weight of the mercury which was forced out. From this, the volume of the mer

diminished in volume, and the index at в was forced towards the bulb at A. The point where it became stationary was marked, and this determined the volume of the air at 0° Centigrade, since the capacity of the bulb was known. The ice was then removed and replaced by water or oil, and the box was heated over a furnace. The air in the bulb expanded, and the index advanced from A towards B. The point where it became stationary was then marked, and at the same time the temperature indicated by the two thermometers D and E was noted, so that the volume of air and its temperature were both ascertained.

Now if in the first instance we suppose that the atmospheric pressure did not vary during the experiment, and neglect the expansion of the glass, which is very small, the total expansion of the air in the apparatus will be found by subtracting the volume which it had at 0° Centigrade from the volume which it had at the end of the experiment. Then, by dividing the remainder by the number of degrees in the final temperature, the expansion corresponding to 1 Centigrade will be deterunits contained in the volume at 0° Centigrade, we obtain the mined; and again dividing this quotient by the number of expansion corresponding to one degree and one unit of volume, that is, the co-efficient of expansion. In the following problems, the corrections for the variations in the pressure of the atmosphere and the expansion of the glass are taken into

account.

Problems on the Expansion of the Gases.-1st. Given v the volume of a gas at 0° Centigrade, and the co-efficient of expansion a, what will be its volume v' at the temperature to Centigrade, the pressure remaining constant. Here, the same reasoning being employed as in the case of linear expansion given in a former lesson, we shall, at once, have v'v+avt; whence v'v (1+at) (1.)

2nd. Given v' the volume of a gas at the temperature to, and the co-efficient of expansion a, what will be its volume v at 0 Centigrade, the pressure remaining constant? Here, from

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cury remaining in the apparatus was deduced, and consequently the volume of the bulb, by proportion, as already shown in the case of the piesometer, p. 105, vol. iv. The bulb and the stem were now filled with dry air, in the following manner: they were first filled with mercury, which was boiled in the bulb in order to dry it; a tube c, filled with drying substances, such as chloride of calcium, was then fixed to the extremity of the stem by means of a cork. Into the stem AB, through the tube, was next introduced a fine platinum wire, by means of which the interior of the tube was agitated; and it was at the same time held in an inclined position, so as to permit the mercury to flow out drop by drop, by slightly shaking the apparatus. The air then entered the bulb A, bubble by bubble, after having been dried by passing through the chloride of calcium. Lastly, there was preserved in the stem AB a small portion of mercury to serve as an index, as The air-thermometer was then placed in a rectangular box made of tin. This box was at first filled with ice, the air was

ahown at B.

VOL. V.

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pressure H. Now by the law of Mariotte, we have vh dried; for the humidity which adhered to its sides is dissipated

VH

1+at whence, v=

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(4.)

5th. Given the volume v' of a glass bulb at t° Centigrade to find its volume v at 0° Centigrade. In solving this problem, we suppose that the bulb will expand for a given variation of temperature, by the same quantity that it would expand were it a mass of glass of the same volume and at the same temperature. If we then represent the co-efficient of the cubic expansion of glass by d, we have according to formula (1.) v=v+ dvtv (1+dt); whence v =

V 1+ di

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In the determination of the co-efficient of the expansion of gases, M. Regnault employed in succession, four different processes. In one set the pressure was constant, and the volume of the gas variable, as in the process of Gay-Lussac; in the other set, the volume remained the same, but the pressure was varied at pleasure. The following is the first process employed by M. Regnault, and in it the pressure remained constant; the same process was employed by M. Dulong and by M. Rudberg. The experiments of M. Regnault are characterised by the greatest care to avoid the possibility of error in the results. His apparatus was composed of a cylindric bulb or reservoir B, fig. 184, of a pretty large capacity, to which is cemented a bent capil

in steam, and the air which filled it each time that a vacuum was made, was dried in its passage through the U-shaped tubes. When this is done, a space of time, say about half-anhour, is allowed for the air to assume the temperature of the steam; the dessiccating tubes are then removed, and the extremity of the capillary tube is hermetically sealed, the height of the barometer at the same instant being noted. The reservoir в is then cooled, and placed in the apparatus shown at fig. 185.

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It is then completely surrounded with ice, so that the air which it contains may be reduced to the temperature of 0° Centigrade; and the extremity of the capillary tube is immer. sed in a cup filled with mercury. When the reservoir Bis at 0° Centigrade, the end b is broken off with a pair of small Fig. 181.

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lary tube. In order to fill this bulb with perfectly dry air, it was arranged, as shown in the figure, in a tin vessel similar to that employed in determining the boiling point of the thermometer; then by means of sheets of caoutchouc, the capillary tube is connected in a series of U-shaped india-rubber tubes, filled with dessiccating substances. These tubes terminate in a small air-pump, by means of which a vacuum is made in the tubes and in the reservoir, whilst the latter is surrounded with steam. The air is then allowed to enter slowly, and a new vacuum is made; and this process is repeated a great number of times; so that at last the air in the reservoir is completely

pincers, and the interior air being condensed, the mercury in the cup enters the capillary tube in consequence of the pres sure of the atmosphere, and rises to a height CG, the pressure of which, added to the elastic force of the air remaining in the apparatus, makes an equilibrium with the atmospheric pres sure. In order to measure the height of the column c G, which may be represented by h, a moveable rod go is lowered until the point o is level with the surface of mercury in the cup; and the difference of the height between the point g and the level of the mercury at a is then measured with the cathetometer. Adding to this difference the length of the part go, which is

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