Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[blocks in formation]

Τοις Αθηναίοις πολλαι ησαν νῆες. Τῳ Διϊ ησαν πολλοί νεῳ,
Οἱ ιχθυες εκ του ύδατος ανακύπτουσιν,
Ο κυβερνήτης την
ναυν ιθύνει. Η ναυς ιθύνεται ύπο του κυβερνητου.
τον Δία και τον Απόλλω,

GREEK-ENGLISH.

Σεβεσθέ

To drink much wine is an evil. Kings have large revenues. In Egypt there was much want of corn. The sea is great. We call a great suffering an Iliad of evils. Croesus had much We willingly yield to soft words. The great gifts of fortune wealth. Often from short pleasure (there) arises great grief. have (are attended by) fear. The manners of many men are mild. Labour greatly assists virtue. Children love their mild fathers and their mild mothers. Have intercourse with mild (gentle) men. The women are gentle. Many call Alexander, the king of the Macedonians, great. ENGLISH-GREEK,

No. XI. Vol. IV. p. 71.-GREEK-ENGLISH. Wantonness begets insult. In meat and drink many are friends, but in a serious undertaking few. Wealth frees men from deficiency and need. Follow nature. Bodily desires (desires springing from the body) occasion wars and insurrections and battles. In states the rulers are guardians of the laws. Ο citizens, abstain from insurrections. O men, long for honourable actions. The natures of mortals are different. From insult many evils arise. The gifts of a bad man have (bring) no profit. Glory and wealth without understanding are not safe possessions. The fruits of figs are sweet. Only the possessions of Virtue are firm. Many towns have walls. The towers of the town are firm. The towers are an ornament to the town.

ENGLISH-GREEK.

Ω παι

Πολλου οίνου απέχου.

Οἱ κακοὶ πολλῳ οινῳ τέρπονται.

Πολύς οινος τους ανθρώπους βλάπτει. Τοις βασιλεύσι μεγάλαι εισι προςοδοι. Η της βασιλείας προςοδος μεγαλη εστιν. Αιγύπτῳ σιτος εστι πολυς. Πολλοις πλουτος μεν εστι πολύς, μικρός δε νοῦς. Των πραέων εθεων ορέγου. Τα των γυναικών εθη πραέα Τοις πράεσι έθεσι εστι κόσμος. Αλεξανδρος, ὁ Μακεδων βασιλευς, πολλακις απαγορεύεται μέγας.

Ο πλούτος την ενδειαν λυει. Εν πόσει και βρώσει έταιρους εχομεν, αλλ' ουκ εν δυςτυχία. Εν τη πόλει εστι ὁ βασιλευς φυλαξ των νόμων. Πειθον, ω νεανια, ταις αρχαις. Ορεγου των καλών πράξεων. Μόνη ή της αρετης κτησις εστι βέβαια. Τη πολεί πολλοί εισι πύργοι. Οἱ καλοι νομοι δοξαν τη πόλει παρεχουσιν. Τη φύσει που. Οἱ στρατιώται περι Φευγε, ω πολιτα, την

της του αστεος σωτηρίας μάχονται, στασιν.

P. 72.-GREEK-ENGLISH.

εστιν.

[blocks in formation]

Women rejoice in ornament. The Greeks worship Zeus (Jupiter), and Poseidon (Neptune), and Apollo, and other Modesty becomes women. gods. Dogs guard the house. The steersman directs the ship. Drops of water hollow out a rock. It is the woman's office to keep the house. It is the part of a good wife to preserve the home. The dice of Zeus always fall well. Dogs occasion men pleasure and profit. The testimonies of witnesses are often incredible. Looms, not public meetings, are the part of wives. Carry the key of the chest, O boy. O Zeus, receive the request of the unfortunate. Castor and Pollux were saviours of ships. Silence gives ornament to every woman. The Ethiopians have black hair. woman, preserve thy home. We comb the hair with a comb. Eacus keeps the keys of Hades.

ENGLISH-GREEK.

IN fig. 36, let A c and EG be parallelograms upon equal bases BC and FG, and between the same parallels A H and BG. The parallelogram A c is equal to the parallelogram EG.

A

Fig. 36. DE

Η

C F

Join B E and on. Because Be is equal to re, (Hyp.) and ra to Εκ (Ι. 34), therefore, nc is equal to EH (Ax. 1). But Bo towards the same parts by the and EH are parallel, and joined straight lines B E and ch. And B straight lines which join the extremities of equal and parallel straight lines towards the same paris, are (I 33) themselves equal and parallel. Therefore the straight lines BE and CH are both equal and parallel. Wherefore BH is a parallelogram (Def. 36). Because the parallelograms Ac and By, are upon the same base B c, and between the same parallels в cand AH, the parallelogram A c is equal (I. 35) to the parallelogram BH, And because the parallelograms GE and HB are upon the same base EH, and between the same parallels GB and HE, the parallelogram EG is equal to the parallelogram BH. Therefore the parallelogram A c is equal (dz. 1) to the parallelogram EG. Therefore, parallelograms upon equal bases, etc. Q. E. D.

EXERCISE I. TO PROPOSITION XXXVI.

If the base of a parallelogram be equal to half the sum of the two parallel sides of a trapezoid, between the same parallels, the parallelogram is equal to the trapezoid.

In figo, let ABCD be a trapezoid, of which the two sides AD and BC are parallel; and let ABLK be a parallelogram having its base BL equal to half the sum of the sides A D and the parallelogram A B L x is equal to the trapezoid A B CD. BC; and let it be between the same parallels A D and BC; then Των γυναικών εστι την οικίαν φυλάττειν, Τας της οικίας Produce BC to E, making CE equal to AD (I. 2); complete κλεῖς κομίζουσι. Αἱ του οίκου κλεῖς τη μητρί προσκομίζονται. [the parallelogram A DEF, and through c and D draw ce and

Κόσμος πρέπει τη γυναίκι. Κόσμος πρέπει ταις γυναιξί.

Dи parallel to AB; then BG, GH, and н F, are parallelograms
(Const.).
Because AF is equal to в E, and CE to AD (Const.), there-
fore Dr is equal to BC (Ax. 3), and the parallelogram BG to
the parallelogram HF (I. 36). Because A a is equal to BC

[blocks in formation]

Produce HA to meet BC and KL in N and M; KB to meet GF in x; and 1 c to meet ED in 0. Then each of the figures HB and Hc is a parallelogram (Const. and Def. 36.)

Because the parallelograms AG and HB are upon the same base, AB, and between the same parallels A B and G H, therefore they are equal (I. 35). For the same reason, the parallelograms AE and Hc are equal, But HA is equal to x B (I. 34), and BK is equal to н A (Const.); therefore x B is equal to в K (Ax. 1). Now, because the parallelograms Hв and Bм are upon equal bases X B and BK, and between the same parallels HM and K X, therefore they are equal (I. 36). But the parallelograms HB and AG were proved equal; therefore the parallelogram A G is equal to the parallelogram вM (Ar. 1). In the same manner it may be shown that the parallelogram A B is equal to the parallelogram CM. Wherefore the parallelograms AG and AE are equal to the parallelograms BM and CM Ex. 2). But the parallelograms BM and cм are together equal to the parallelogram BL; therefore the parallelograms Wherefore, the parallelograms described on any, ete. Q.E.D. AG and AE are together equal to the parallelogram BL (Ax. 1).

PROPOSITION XXXVII. THEOREM.

(I. 34), and therefore to DF (Ar. 1), it is also equal to (1. 34, and Ax. 1); but A D is equal to CE (Const.); therefore G D is equal to CH, and the triangle & CD to the triangle DCH (I. 34); now, adding the triangle & CD to the parallelogram BG, and the triangle DCH to the parallelogram HF, the trapezoid ABCD is equal to the trapezoid D CEF (Az. 2), and each is half of the parallelogram A B EF; but the parallelogram Triangles upon the same base, and between the same parallels, are ABLK is equal to the parallelogram K LEF (I. 36) and each is half of of the parallelogram A BEF; therefore, the parallelogram A BLK is equal to the trapezoid A B C D (4, 1). Where-base BC, and between the same parallels AD and BC. In fig. 37, let the triangles A B C and D BC be upon the same fore, if the base of a parallelogram, etc.* Q. E. D.

[merged small][ocr errors][merged small][merged small][merged small][merged small]

equal to one another.

triangle A B C is equal to the triangle D B C.
Produce A D both ways to the E
points E and F. Through B draw
BE parallel to CA (I. 31), and
through c draw or parallel to
BD. Then each of the figures
EC and BP, is a parallelogram
(Def. 36).

B

The

Fig. 37.

[blocks in formation]

The parallelograms EC and Br are equal (I. 35), because they are upon the same base B C, and between the same parallels B C and E F. But the triangle ABC is half of the parallelogram Ec (I 34), because the diagonal A B bisects it; also, the triangle DBC is half of the parallelogram BF, because the diagonal DC bisects it; and the halves of equal things are equal (Ax. 7). Therefore the triangle ABC is equal to the triangle. Wherefore, triangles, etc. Q. E. D.

DB C.

EXERCISE TO PROPOSITION XXXVII.

To describe a triangle equal to any given rectilineal figure. N.B. In solving this problem, the learner may begin with a parallelogram; then proceed to a trapezium, a pentagon, a hexagon, and so on. He will then ultimately find that, if a figure had a thousand sides, he could reduce it, by degrees, to an equivalent triangle.

In fig. 9, let ABCDEFG be any rectilineal figure; it is required to describe a triangle equal to it.

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

described. Also, let their exterior sides ED and GF be pro-
duced till they meet in the point H (Ax. 12), and join HA.
Through the point в draw BK equal and parallel to HA (I. 31
and 3). Complete the parallelogram B L.
Then the parallelo-

H A

Take any side A B and produce it indefinitely in the direction AI; join AF, and through G, draw a H parallel to AP (I. 31), grams AG and AE together are equal to the parallelogram join FH. Then the rectilineal figure HBCDEF is equal to the

BL.

Solved by WARIN (East Dereham); E. J. BREMNER (Carlisle); J. H. EASTWOOD (Middleton); QUINTIN PRINGLE (Glasgow); and others.

rectilineal figure A B C D E F G.

* Solved by WARIN (East Dereham); J. H. EASTWOOD (Middleton); J. BREMNER (Carlisle); and others

Because the triangles AGF and AHF are upon the same base AF, and between the same parallels GH and A F, therefore the triangle AHF is equal to the triangle AGF (I. 37). To these equals, add the rectilineal figure ABCDEF, and the whole HBCDEF is equal to the whole ABCDEFG (4x. 2); that is, the former figure is equal to the latter, but it has one side fewer than it. Again, join HB, and through F, draw FI parallel to HE (I. 31); join EI. Then the rectilineal figure IBCDE is equal to the rectilineal figure H BCDEF.

bisects it. And the triangle DEF is the half of the parallelogram E H, because the diagonal BF bisects it. But the halves of equal things are equal (Ax. 7). Therefore the triangle A B C is equal to the triangle DEF. Wherefore, triangles.upon equal bases, etc. Q. E D.

Corollary 1.-The straight line drawn from any angle of a triangle bisecting the opposite side, bisects the triangle. Corollary 2.-If two triangles have two sides of the one equal to two sides of the other, each to each, and the angle contained by the two sides of the one, the supplement of the angle contained by the two sides of the other, these triangles are equal,

EXERCISE I. TO PROPOSITION XXXVIII.

Because the triangles HFE and HIE are upon the same base HE, and between the same parallels FI and H E, therefore the triangle HIE is equal to the triangle HFE (I. 37). To these equals, add the rectilineal figure HB CD E, and the whole IB C DE is equal to the whole HBCDEF; that is, the former figure To bisect a triangle by drawing a straight line through any point in is equal to the latter, but it has one side fewer than it. Now the rectilineal figure HBCDEF was proved equal to the rectilineal figure ABCDEFG; therefore, the rectilineal figure IBC DE is equal to the rectilineal figure ABCDEFG (Ax. 1); and

the former has two sides fewer than the latter.

In like manner, by joining ID, and drawing through E, a parallel to ID, meeting AB produced as before, a rectilineal figure may be found equal to the rectilineal figure ABCDEFG, and having three sides fewer than the latter; and so on, may the operation be continued, until a figure having only three sides, that is, a triangle, may be found, which shall be equal to the given rectilineal ABCDEFG. Q. E. F. *

Scholium.-This proposition is of extensive use in the practice of land-surveying; for whatever may be the form of the field to be measured, if the lengths of all the sides be correctly ascertained, as well as their relative positions to one another, that is, the angles which every two adjacent sides form between them; then, by laying down a plan of the field upon paper, so that in the plan, the length of every side shall have to the length of every other the same relative proportions which they have in the field, the angles being laid down carefully with the protractor, and the sides measured to nicety by means of the plane scale, the figure of the field may be reduced to that of a triangle, as shown in the proposition. When this is done, the area of the triangle can be found by measuring its base and perpendicular from the same plane scale used in laying down the sides, and applying the following Rule: Multiply the length of the base in numbers, by the length of the perpendicular in numbers, and divide the product by 2; the quotient will be the area of the triangle in square units, these being the squares of the units employed in measuring the lengths of the sides.

Definitions.-The base of a triangle is any one of its sides, to which a straight line is drawn at right angles, from the vertex of the angle opposite to that side. This angle, in reference to the base, is called the vertical angle; and the straight line drawn at right angles to the base is called the perpendicular of the triangle. When one of the angles at the base of the triangle is obtuse, the perpendicular will fall without the triangle, upon the base produced; otherwise (except in the case of the right-angled triangle having its right angle at the base), the perpendicular will fall within the triangle, upon the base itself.

PROPOSITION XXXVIII. THEOREM.

one of its sides.

In fig. r, let A B C be a triangle, and D a point in one of its sides BC; it is required to bisect the triangle ABC, by draw ing a straight line through D.

[merged small][merged small][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small]
[blocks in formation]

But

If the point D were in the middle of the base PC, as at E, then by joining EA, the thing required is done (I. 38, Cor. 1); that is, the triangle ABI is equal to the triangle A EC. if the point D be not in the middle of the base, join DA; bisect BC in E (I. 9), join E A, and through the point E, draw EF parallel to AD (I. 31); and join DF. Then DF bisects the triangle A B C.

Then each of the

Because the two triangles A FE and DFE are upon the same base FE, and between the same parallels AD and FE, the triangle AFE is equal to the triangle DFE (I. 37). To these equals, add the triangle A RE, and the whole triangle ABB is equal to the whole triangle FBD is half of the triangle ABC (I. 38, Cor. 1); therefore the triangle FBD is also half of the triangle ABC (Ax. 1); and the triangle ABC is bisected by the straight line DF. Q. E. F.*

The parallelograms GC and E H are equal to one another (I. 36), because they are upon equal bases BC and EF, and between the same parallels BF and GH. The triangle ABC is half of the parallelogram G C (I. 34), because the diagonal AB

Solved by J. H. EASTWOOD (Middleton); 'Q. PRINGLE (Glasgow); WARIN (East Dereham); and others.

But, the triangle ABE

[blocks in formation]

Draw the diagonal AD, and let it meet the diagonal Bc in F. Because AF is equal to F D (Exercise 2, Prop. xxxiv.) there. fore the triangle AFC is equal to the triangle DFC, and the triangle AFE to the triangle DFE (I. 38, Cor. 1); therefore the whole triangle AC is equal to the whole triangle DCB (4x. 2). But the triangle ABC is equal to the triangle Dnc (I. 34); therefore the remaining triangle A B E is equal to the remain

Solved by J. H. EASTWOOD (Middleton); J. JENKINS (Pembroke Dock); WARIN (East Dereham); E. J. BREMNER (Carlisle); and others.

ing triangle DB E (Ax. 3). Therefore the straight lines, etc. | reference to Prop. iv. Book I. instead of to Prop. xxvi. Book Q. E. D.*

I. is a piece of unnecessary refinement; because Prop. iv.
Book I. is referred to in Prop. xxvi. in order to establish the
equality of the triangles, and therefore the condition required
by the former proposition is fulfilled in the demonstration of
Prop. xxvi. It is therefore fair to consider that the require-
ments of Prop. iv. are included in the demonstration of Prop.
XXVI. and need not be referred to again.
which we have given on the propositions from Prop. xxxiv. to
Some additional exercises might have been added to those
Prop. xxxviii., such as the following, which we submit to our
students by way of stirring them up a little to the applica-
tion of Euclid.

1. If the diagonals of a parallelogram be equal, it is right-
angled.
2. The diagonals of a rhombus bisect its angles.
3. If the diagonals of a quadrilateral figure bisect its angles,
it is a rhombus.

4. Those parallelograms which have any two adjacent sides and their included angle equal, are equal in all respects. 5. If the opposite sides of a parallelogram be divided into the same number of equal parts, and if the corresponding points of division in each be joined by straight lines, the parallelogram will be divided into as many equal parallelograms as there are equal parts in each of the opposite sides which have been so divided.

6. If the base of a triangle be divided into any number of equal parts, and straight lines be drawn from the vertex to the several points of division, the triangle will be divided into the same number of equal parts.

FRENCH READING S.-No. XVI.

CHARLES I.

COURAGE ET GRANDEUR DANS L'IN FORTUNE.

General Scholium to this Lesson.-In our additions to Prop. xxxii. we forgot to advert to the demonstration of Pappus, which is effected without producing the base в C, see fig. 32, No. 1, p. 122, vol. v. Instead of producing EC, Pappus draws through the vertex a, a straight line parallel to the base B C, and proves, at once, that the three angles A B C, B C A and BAC are equal to two right angles, by means of Cor. 1, to Prop. xiii.; to demonstrate this will prove a useful and easy exercise for our students. The demonstration of Euclid's Cor. I. viz. "that 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," may be conducted without assuming a point F, within the figure, see fig. 32, No. 2, p. 122, vol. v., and drawing straight lines from it, to each of the angles. Thus, by drawing from any angular point a, straight lines to each of the other angles, it is plain that the figure will be divided into as many triangles as the figure has sides, wanting two; because the straight lines drawn to the two adjacent points B and E, will coincide with the two sides A B and A E, and no triangles will be formed with these sides; but triangles will be formed with each of the other sides. Now as all the angles of each triangle thus formed, are equal to two right angles, all the angles of all these triangles are equal to twice as many right angles as there are sides of the figure, excepting two sides; and had triangles been formed on these two sides, all their angles would have been equal to four right angles. Therefore, all the angles of all the triangles thus formed, together with four right angles, are equal to twice as many right angles as the figure has sides. But all the angles of all the triangles thus formed are equal to ail the angles of the figure (Const.); 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 (4x. 1). Q. E. D. The demonstration of Euclid's Cor. II. viz. "that 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," may be conducted on a different principle from that adopted by him, and perhaps in some respects foreign to the Elements of the Greek Geometry, but still equally convincing. Thus suppose that the figure in p. 123, Nous n'avons pas l'intention de justifier toute la conduite rol. v. fig. 32, No. 3, were replaced by a smaller one which de Charles premier, roi d'Angleterre, époux de Henriettehad its angles exactly the same, and this by a smaller still Marie de France, fille de Henri quatre. Il avait commis having the same angles; and so on, till the interior of the bien des fautes; sa faiblesse avait été jusqu'à lui faire signer figure was reduced to a point; the straight lines which form l'arrêt de mort d'un ministre qui s'était perdu à le servir. the exterior angles of the figure would not be altered in their Ses perpétuelles hésitations, ses incertitudes de caractère3 direction, whatever might be their number; but "all the n'avaient su ni arrêter ni prévenir la guerre civile; et son angles made by any number of straight lines meeting in a are together equal to four right angles;" therefore all courage chevaleresque dans la bataille, l'excellence et la géthe exterior angles of any rectilineal figure, made as described, nérosité de son cœur en de nombreuses circonstances n'auare together equal to four right angles. Indeed it might be raient point suffi à effacer les taches de son déplorable considered enough to say, that as the space within the recti- règne. Les derniers jours de son orageuse vie, et le voile lineal figure, being limited on all sides, may be considered as sanglant qui se tira sur sa dernière heure, purent seuls faire amere point compared with infinite space, therefore all the oublier le monarque inhabile, pour ne plus laisser voir que angles made by the straight lines produced about this point, le grand et courageux martyr.5 are, as before, equal to four right angles. Q. E. D. Prisonnier depuis deux ans, Charles ne pouvait cependant The remarks made by Dr. Simson in reference to his demon croire qu'on osat en venir jusqu'à lui faire réellement son stration of Prop. xxxv. which we have given at p. 183, vol. v. procès; ce qu'il craignait plus sérieusement, c'était un asmust not be omitted; they are as follows: "The demonstra-sassinat nocturne. Harrison, colonel des troupes du Pariewhich is used in it was followed, there would be three cases an of this proposition is changed, because, if the method ment, le tira d'erreur en lui disant que sa mort serait aussi to be peu obscure que l'est le soleil en plein midi. En effet, le 28 than the Arabic; for in the Elements, no case of a proposition décembre, 16.8, une haute cour de justice, composée de a different demonstration ought to be omitted. membres choisis par Cromwell, fut instituée pour faire le On this account we have chosen the method which Mons. procès du roi. Ce procès commença au mois de janvier Eements, page 21, and which afterwards Mr. [Thomas] craignait une réaction en faveur d'un prince si malheureux. Lairault has given, the first of any, as far as I know, in his suivant, et fut conduit avec d'autant plus de célérité, qu'on But whereas Mr. Sinipson Pendant tout le cours de débats, Charles déploya une fermeté 12 qui ne fit que croître avec l'imminence du danger, et ce fut sans donner aucune marque apparente d'émotion, 13 qu'il entendit la lecture de l'arrêt qui le condamnait comme tyran, comme traitre et ennemi public, à avoir lu teten

that

requires

Sapon gives in his, page 32.

makes use

of Prop. xxvi., Book I. from which the equality of two triangles does not immediately follow, because to prove that, the 4th of Book I. must likewise be made use of, as seen in the very same case in the 34th Prop. Book I., On these remarks of Dr. Simson, we would observe that the U the thought better to make use only of the 4th of Book L."

may

tranchée sur le billot.

SECTION I.

On avait laissé à Charles 13 trois jours pour se préparer à la mort. ed by J. H. EASTWOOD (Middleton); E. J. BREMNER (Carlisle); même qui avait été choisie pour son supplice. Ce fut là Il les passa à Whitehall, en face de la place 16

WAN (East Dereham); and others.

[blocks in formation]

15. Quel délai avait-on accordé à Charles?

16. Où passa-t-il ces trois jours?

17. Que demanda le roi? 18. Quel âge avait le duc de Glocester ?

¦ 19. De quoi les enfants ne se doutaient-ils pas?

20. Que ne soupçonnaient-ils pas?

NOTES AND REFERENCES.-a. from commettre; L. part ii., p. 82; also L. S. 41, R. 8.-b. perdu, ruined.-c. L. part ii., § 138, R. (1)-d. osit, would dare.-e. from craindre; L. part ii., p. 84.-f. avoir la tête tranchée, to be beheaded.

SECTION II.

-Pour qui donc tous ces hommes construisent-ils ce trône tendu de noir? demanda-t-il naïvement.

Le roi tressaille à cette question,18 non pour lui-même, mais pour ses enfants; il sent que le moment est venu de déclarer toute la vérité.

-C'est pour votre père, mes enfants, dit-il; ce n'est pas un trône, c'est un échafaud.

Elisabeth jette un cri perçant et s'évanouit; elle avait enfin 20 tout compris.

Les cruels, dit Charles, clouer 21 ainsi l'échafaud du père à la vue de ses enfants!

COLLOQUIAL EXERCISE.

1. Quel soin avait-on laissé au
roi?

2. Comment Charles accueillit-
il ses enfants?

3. Que crurent d'abord les en-
fants?

4. Pourquoi pleurait-il ?
5. Que fit-il, après le premier
mouvement de douleur?

On avait laissé au roi le soin cruel d'éclairer ses enfants sur son sort prochain. Quand ils furent introduits," son courage parut l'abandonner un moment, et il les serra sur son cœur en pleurant. Elisabeth et le petit duc crurent que c'était du bonheur de les revoir. Hélas! c'était de l'idée de s'en séparer bientôt pour toujours, au moins sur la terre. Mais ce premier mouvement passé, il rassembla toutes ses forces, et prenant une des mains de sa fille dans sa main, appuyant sa joue contre la sienne, asseyant le petit duc de Glocester sur ses genoux, il leur parla de la mort en termes un peu vagues et généraux, essaya de leur faire pressentir une séparation longue et qui ne devait avoir que le ciel pour point de réunion. Elisabeth essayait de démêler quelque chose dans cette obscurité; elle mouillait de pleurs la main et le visage de son père; mais, après la captivité, son imagination, à force de se tourmenter, ne se creait encore aucune image plus funeste que l'exil. Pour le petit due, il regardait son père avec de grands yeux fixes et étonnés ; il réfléchissait profondément, pour un enfant de son âge, et ces paroles, qui émurente profondément le roi, sortirent de sa bouche:

-Est-ce qu'il est un plus grand malheur" que de ne pas embrasser sa mère et de vivre en prison loin de son père, comme nous faisons depuis si long-temps?

6. Que fit-il encore après avoir
placé le petit garçon sur ses
genoux?

7. Qu'essaya-t-il de leur faire
pressentir?

Oui, mon fils, lui répondit le roi, il est un malheur plus 12 grand.

8. Qu'est-ce qu'Elisabeth essa-
yait de faire?

Elisabeth ne comprenant pas encore,13 et son frère encore moins.

9. Soupçonnait-elle la vérité?
10. Que faisait alors le petit duc?

-Oh! dit-elle, est-ce que l'on vous emmènera14 bien loin, sans nous ?

-Bien loin, répondit le roi.

11. Que dit-il à son père? 12. Que lui répondit le roi ? 13. Elisabeth comprenait-elle le vrai sens des paroles de son père ?

14. Que dit-elle alors? 15. Qu'entendit-on alors, du côté de la fenêtre ? 16. Où était allé le petit due? 17. Que demanda-t-il naïve

ment?

18. Quel effet la question eut elle sur le roi?

19. Que répondit-il? 20. Elisabeth avait-elle enfin compris la vérité?

21. Que dit alors Charles?

NOTES AND REFERENCES.- -a. L. S. 45, R. 2.-b. from asseoir; L. part ii., p. 78.— -c. L. S. 31, R. 5.-d. pour, as to.-e. from mouvoir; L. part ii., p. 88.-f. est-ce qu'il est, is there; L. S. 21, R. 3, 4.-g. L. part ii, § 49, R. (6).—h. from construire; L. part ii., p. 81.

Pendant ce temps, un grand bruit de marteaux et d'ouvriers s'étant fait entendre du côté de la fenêtre de la chambre ou avait lieu cette triste entrevue, le petit duc était descendu machinalement des genoux de son père et était allé à la fenêtre.

SECTION III.

Le petit duc ne s'expliquait pas encore les choses; il fut épouvanté du eri et de l'évanouissement de sa sœur, mais, sans y rien comprendre, il adressa encore au roi plusieurs questions ingénues et déchirantes, entre autres, il lui demanda si ce n'était pas pour le rétablir dans son pouvoir que ses sujets dressaient cet appareil, et si lui, le petit duc, il serait de la fete? C'étaient mille morts pour une, que le père souffrait à l'avance.

O mon fils, dit-il, après avoir rappelé sa fille de son évanouissement, et en replaçant l'enfant sur ses genoux, mon fils, écoute bien cela, il vont tuer ton père.

6

L'enfant regarda de nouveau fixement le roi; puis un jour immense et terrible se faisant tout à coup dans son jeune esprit, il s'écria avec une force surprenante pour son age:

-Vous tuer, vous, jamais! je me ferais plutôt hacher en morceaux que de le souffrir.

-Le roi émerveillé de cette parole courageuse de son fils, l'embrassa par trois fois avec effusion, et s'écria: -C'est là un digne enfant que m'a donné la fille de Henri IV.!

Un des officiers commis à la garde de Charles,' ayant entendu le mot du petit duc, dit de son côté :

de

-Voilà un enfant dangereux et dont il sera 10 à propos se défaire, avant que la force de son corps égale celle de son caractère.

Le plus fort étant fait, le roi fut plus à l'aise pour adresser ses dernières recommandations à ses enfants.

--Souviens-toi, dit-il à son fils, que la couronne d'Angleterre appartient après moi à ton frère ainé, à qui tu dois obéissance. Peut-être qu'après ma mort les Anglais voudront te donner la couronne; promets-moi de ne pas cepter.13

[blocks in formation]
« ΠροηγούμενηΣυνέχεια »