Indeterminate Preterite. darà, he will give darémo, we will give désti, thou gavest daréte, you will give diéde, détte or diè, he gave dardnno, they will give INFINITIVE Mood. démmo, we gave Conditional Present. déste, you gave Daréi, I should or would diédero, déttero, diérono or diés give Present: andare, to go Past : essere andato, to have dono, * they gave darésti, thou wouldst give or be gone Future. darébbe, he would give Present Gerund : andando, Past Gerund : essendo andato, darémmo, we would give going having, or being gone Darò, I shall or will give daréste, you would give darébbero, they would give IMPERATIVE MOOD, [No First Person). diamo, let us give INDICATIVE Mood. Dà, give (thou) dáte, give (ye or you) dia or déa, let him give diano or dieno, let them give Vado or vo, I go anddste, you went tái, thou goest andárono, they went SUBJUNCTIVE MOOD. ra, he goes Future. Imperfect. andiamo, we go Che dia, that I may give Che déssi, that I might give che dia or dii, that thou mayst | che déssi, that thou mightst oanno, they go andrà, he will go give give che dia, that he may give che desse, that he might give che diámo, that we may give Andava, I was going che déssimo, that we might give che dicite, that you may give andávi, thou wast going andranno, they will go che déste, that you might give che diano, dieno, déano, that che dessero, that they might andáva, he was going Conditional Present. give andavamo, we were going Andrei, I should or would andavite, you were going So conjugatego andrivano, they were going andresti, thou wouldst go Ridáre, to give again. Addáre or addársi, to apply one's self. Indeterminate Preterite. andrebbe, he would go Andai, I went andremmo, he would go III. andásti, thou wentest andréste, you would go Fare, to make. INFINITIVE MOOD. [No First Person.] andiamo, let us go Simple Tenses. Compound Tenses. Va, go (thou) andate, go (ye or you) sáda, let him go vádano, let them go Present: fáre, to make Past : avére fatto, to have made Present Gerund : facéndo, Past Gerund : avendo fatto, Imperfect. having made Past Participle : fátto, made che odda or vádi, that thou che andassi, that thou mightst mayst go go INDICATIVE MOOD. che odda, that he may go che andusse, that he might go che andiamo, that we may go che andassimo, that we mightgo Present. féce, fè, or féo, he made che andiate, that you may go che andúste, that you might go Fo or fáccio, I make facemmo, we made che oddano, that they may go che andassero, that they might fái, thou makest facéste, you made fécero or fénno, they made facciamo, we make So conjugate Future, Faro, I shall or will make fanno, they make farái, thou wilt make Imperfect. farà, he will make II. Faceva, facéa, fè, I was making | faréte, you will make farémo, we will make Dáre, to give. facévi, thou wast making INFINITIVE MOOD. faceva or facea , he was making farinno, they will make Conditional Present. make Present: dáre, to give Past: avere dito, to have given facevano or faceano, they were making farésti, thou wouldst make Present Gerund: dándo, giv- Past Gerund: avendo dato, ing farebbe, he would make having given Indeterminate Preterite. farémmo, we would make Past Participle : dato, given Fici or féi, I made faréste, you would make facesti, thou madest 'farebbero, they would make IMPERATIVB MOOD. Imperfect. [No First Person.) facciamo, let us make fåte, make (ye or you) da, he gives dura, he was giving facciano, let them make diumo, we give daváno, we were giving dáte, you give davate, you were giving alkinno, they give dávano, they were giving • Diér, diéro, dénno, are used in poetry, mayst make 1 SUBJUNCTIVE MOOD. IRREGULAR VERBS OF THE SECOND CONJUGATION. Imperfect. Verbs ending in ere are of two sorts. The first have their Che faccia, that I may make Che facéssi, that I might make infinitives long, such as bére, cadere, etc.; the second short, che faccia or facci, that thou che facessi , that thou mightst such as assórbere, conoscere, etc. 1st. Irregular verbs ending in ére long. 1. make Bére, to drink. INFINITIVE Mood. make make So conjugate Simple Tenses. Compound Tenses. Present: bére, to drink Past : avere beúto, to hare drunk Sfare, to undo Sopraffare, to ask too much Present Gerund : bevéndo, Past Gerund : avendo beute, drinking having drunk Liquefare, to melt Strafore, to do too much Past Participle: beúto, drunk INDICATIVE Mood, Present. bévre, he drank Stáre, to stand. Béo, I drink beémmo, we drank beéste, you drank béi, thou drinkest INFINITIVE Mood. bée, he drinks bévvero, they drank Future. beéte, you drink berò, I shall or will drink berdi, thou wilt drink Imperfect. berà, he will drink berémo, we will drink beréte, you will drink beévi, thou wast drinking berdnno, they will drink beevamo, we were drinking Conditional Present. beeváte, you were drinking Beréi, I should or would drink Present. stétte, he stood beévano, they were drinking berésti, thou wouldst drink Sto, I stand stémmo, we stood berébbe, he would drink stái, thou standest Indeterminate Preterite. stéste, you stood berémmo, we would drink ata, he stands stéttero, they stood Bévvi, I drank beréste, you would drink Future. berébbero, they would drink státe, you stand Staro, I shall or will stand IMPERATIVE MOOD. [No First Person. beiamo, let us drink beéte, drink (ye or you) béano, let them drink staránno, they will stand stavámo, we were standing Conditional Present. SUBJUNCTIVE MOOD. staváte, you were standing Staré, I should or would stand Present. drink che béa, that thou mayst drink | che beesse, that he might drink starémmo, we would stand che beéssimo, that we might che béa, that he may drink drink che heidmo, that we may drink starebbero, they would stand che beiáte, that you may drink che beeste, that you might drink che béano, that they may drink IMPERATIVE MOOD. che beéssero, that they might Imperfect drink Che beéssi, that I might drink So conjugate- Imbére, to imbibe | Ribére, to drink again Strabére, to drink hard. The Italians prefer the regular verb bévere. II. stand Cadére, to fall. INFINITIVE Mood. Simple Tenses. Compound Tenses. stand Present: cadére, to fall Past: éssere caduto, to have fallen Present Gerund: cadendo, fall- Past Gerund: essendo caduto, ing having fallen Past Participle: cadúlo, fallen che betinis that thou mightst may stand A D From the point a draw a B perpendicular to BC (I. 12). INDICATIVE MOOD. Then, because the angle A EB is a right angle (Cons.), therefore Present. Future. the square of A B is equal to the squares of B E and zä together (I. 47). For the same reason the square of a c is equal to the squares of cádi, thou fallest fall code, he falls cadrdi or caderái, thou wilt fall 4 E and E c together. Therefore, adding equals to equals (I. Az. 2), the squares of . , cadiómo or caggidmo, we fall cadrà or caderà, he will fall A B and a c together are equal to the squares of BB and so cadéte, you fall cadremo or caderémo, we will together with twice the square of Ba. cadono or cággiono, they fall fall But because Bc is divided into two equal parts in D and two Imperfect. cadréte or caderéte, you will fall unequal parts in E, therefore the squares of BB and Bo are Cadeva or cadéa, I was falling cadranno or caderánno, they together equal to twice the square of BD together with twice the will fall cadévi, thou wast falling square of D E (II. 9). cadeva or cadea, he was falling Conditional Present. Therefore the squares of AB and AC together are equal to cadevamo, we were falling Cadréi, or caderéi, cadería, ca- twice the squares of B D, D B, and B A together. cadeváte, you were falling dría, I should or would But the squares of D B and BA are together equal to the square cadévano or cadéano, they were fall of AD (I. 47), and the doubles of these equals are equal. Therefalling cadrésti or caderésti, thou fore twice the squares of D E and BA are equal to twice the square Indeterminate Preterite. wouldst fall of AD. Therefore the squares of AB and a c together are equal Caddi, cadetti, or cadéi, I fell cadrébbe or caderebbe, caderia, to twice the squares of BD and Da together. Q.E.D. cadésti, thou fellest cadria, he would fall Prop. B. Theorem. The squares of the two diagonals of a cadde, cadeo, cadétte, or cadè, cadremmo 'or caderemmo, we parallelogram are together equal to the squares of its four sides. he fell would fall cadémmo, we fell cadréste or caderéste, you would fall riano, they would fall cadéte, fall (ye or you) Let ABCD be any parallelogram of which the diagonals are AC and B D cutting each other in .. Then the equares of a c and Present. Imperfect. BD together are equal to the squares of AB, BC, CD, and DA together. By Exercise 2 to Proposition XXXIV. of the first book, Che cada or cággia, that I may che cadéssi, that I might fall fall che cadéssi, that thou mightst of which the side 8 D is bisected in , therefore, by the last pro the diagonals bisect each other. And because ABD is a triangle che cada or caggia, that thou fall mayst fall che cadesse that he might fall position, the squares of 13 and ap together are equal to twice che cada or caggia, that he may che cadessimo, that we might the squares of B 2 and A B together. For the same reason the squares of BC and CD are together fall fall che cadiamo or caggiamo, that che cadéste, that you might adding equals to equals, the squares of a B, BC, CD, and DA equal to twice the squares of B E and 2 c together. Therefore, we may fall fall together are equal to four times the square of Be together with fall che coidano or caggiano, that But A e is equal to e c, because the diagonals bisect each other, and therefore the square of AE is equal to the square of 2C they may fall Therefore the squares of AB, BC, CD, and Da together are equal So conjugate to four times the squares of B and as together, But the square of any straight line is equal to four times the square of half the line (II. 4, Cor. 2). Therefore the square of A is equal to four times the square of A E. Therefore the square of . B, BC, CD, and D A are together equal to the squares of A c and B D together. Q.E.D. Prop. c. Theorem. The squares of the four sides of a traSOLUTION OF EXERCISES TO THE SECOND pezium are together equal to the squares of its two diagonals and BOOK OF EUCLID. four times the square of the straight line which joins the points of the bisection of the diagonals. D B may fall E B B D D Е с Let A B C D be any trapezium whose diagonals a C and B D are Let A B C be any triangle having one of its sides B C bisected bisected in the points F and B. Join E f. Then the equares of A B, in D. Join a D. Then the squares of AB and Ac are together BC, CD, and D A are together equal to the squares of a cand BD equal to twice the squares of B D and A D together. together with four times the square of E F. H E А G B M E C. B Join Ba and Ac. Then, because ABD is a triangle whose the rectangle contained by the whole line thus produced, and the base is bisected in E, and BA is drawn, therefore the squares of part produced, may be equal to a given square. A B and A D together are equal to twice the squares of BE and E A together (II. Prop. A). For the same reason the squares of BC and CD are together equal to twice the squares of BE and EC together. Therefore, adding equals to equals, the squares of A B, BD, CD, and D A, are together equal to four times the square of BE C together with twice the squares of E A and E c together. But the square of BD is equal to four times the square of BE (II. 4, Cor. 2). Therefore the squares of A B, BC, CD, and D A are together equal to the square of B D together with twice the squares of Ea and Again, because EAC is a triangle, of which the base ao is N bisected in e (Hyp.), therefore the squares of EA and ec are together equal to twice the squares of cr and EF together Let A B be the given straight line, and C D E F the given square. (II. Prop. A), And the doubles of these equals are equal; therefore the the whole line thus produced, and the part of it produced may be It is required to produce AB so that the rectangle contained by doubles of the squares of EA and Ec are together equal to four equal to the given square CD e F. times the squares of CF and EP together. From the point B in the straight line A B draw Bu at right Therefore the squares of a B, BC, CD, and D A are together equal angles to A B (I. 11) and equal to CD (I. 3) one of the sides of the to the square of BD together with four times the squares of CP given square CD e F. and EF. Bisect a B in the point G and join G H. But four times the square of cp is equal to the square of AC With centre G and distance Gh describe the circle umn. ни (II. 4, Cor. 2). Produce a B to meet the circumference in L. Then the rectTherefore the squares of A B, BC, CD, and D A are together angle A L, L B is equal to the square CDEF. equal to the squares of B D and á c together with four times the Because Bu is equal to co (Cons.), therefore the square of square of Ep. Q.E.D. Bh is equal to the square of CD, that is the square C D E F. Prop. D. Problem. To divide a given straight line into two Again, because G is the centre of the circle xy x, therefore G I parts, 80 that the rectangle contained by its segments shall be is equal to G L (I. Def. 15) and therefore the square of Gu is equal to a given square, not greater than the square of half the equal to the square of L. given straig bt line. Again, because I B G is a right angled triangle having the right angle u BG (Cons.), therefore the square of G u is equal to the H squares of G B and B 1 together (I. 47). But the square of G H has been above shown to be equal to the square of G i. Therefore (I. Ax. 1) the square of G L is equal to the squares of G B and BH together. But the square of G L is equal to the square of G B together with the rectangle A L, L B (II. 6). Therefore (I. Ax. 1) the squares of G B and rk are together equal to the square of G B together with the rectangle A L, L B. From each of these equals take away the common square of & B, K then (I. Ax. 3) the rectangle a L, LB is equal to the square of Let A B be the given straight line and CDEF the given square But the square of Ba is equal to the square CD E P(Cons.). Therenot greater than the square of half the given line AB. It is fore the rectangle A L, LB is equal to the square CDEP(I. As. 1). required to divide A B into two parts such that the rectangle Q.E.F.. contained by them may be equal to the given square C D E F. Bisect a B in G (I. 10). From a draw gx at right angles to Joan BURROUGH: The POPULAR EDUCATOR has been published in the United States. We cannot promise to act upon your suggestion. With centre 3 and distance is describe the circle x K L, GEORGE YOUNG will find our Lessons in Italian answer his purpose. cutting a B in 2. The straight line A B is divided in M, so that UN FRANÇAIS: It is doubtful whether we shall be able to find room for the rectangle Au, x B is equal to the square CD EF. any lessons in Spanish, but we should be glad if it were possible. Join . Then because a B is divided into two equal parts in G, cator" for articles on ancient history. G. LAXBMAN is referred to the two volumes of the "Historical Edu. The most important parts of and two unequal parts in y, therefore the square of aB is equal Arithmetic have already been given in these pages. If we can find room for to the square of G x together with the rectangle ax, MB. further lessons, we wil; but if not, we must reler to “Cassell's Arithmetic” for any additional information that may be desired. But G B is equal to 11, by construction, and uk is equal to BETH: Your questions are unsuitable for discussion in the " POPULAR XX (I. Def, 15). EDOCATOR." We have no wish to engage in theological controversy. Therefore (I. Ax. 1) GB is equal to 1 m, and the square of G B DesUNT CBTERA: The promise will be fulfilled at the commencement of to the square of us. SIMPLICITAS (Wemyos, Fireshire) has solved the first thirty-two of the Therefore the square of xx is equal to the square of GM Second Centenary of Algebraical Problems; D. HORSBY (Driffield) the first together with the rectangle a M, MB. fifty-five, with the exception of Nos. 7, 12, 35, 37, 39, 45, and 54; GEORGE But the square of ax is equal to the squares of a G and GM And og. Fenchurch-street) the first thirty-two, except Nos. 16, 23, 20, WILD (Dalton-on-Tees) the whole of the second portion from No.?3 to 70; together (I. 47). 28, 30, and 31, besides Nos. 46, 41, and 51 of the second portion. Therefore (I. Ax, 1) the squares of 1 G and ax are together D. BORNBY (Driffieid): The following is George Wild's solution of Proequal to the square of a'r together with the rectangle blem 49:A M, XB. 5 9 Days Days : : 10 : 18 From each of these equals take the square of Gm, then the 9 rectangle a x, x B is equal to the square of 1 G (I. Ax. 3). But the number of days in which A can do the whole. Therefore he would the square of a G is equal to the square CDEP (Cons.). Therefore the rectangle a M, M B is equal to the square CDEP reyuire 8 days to do the remaining (I. Az, 1). Q.E.F. Prog. E. Problem. To produce a given straight line, so that The above five exercises were soived by J. H. EASTWOOD (Middleton). B D E вн. EP. A next year. 9 of the work alone. + Let I be the number of days in which B could do this part of the work Now ready, price 93. strongly bound, CASSELL'S GERMAN PRONOUNCING DICTIONARY In Two Parts :-1. German and English; 2. English and German. In one large handsome Octavo Volume. The German-English Division, price 4 9 Days Days 55. in paper covers, or 53. 6d. neat cloth; the English-German Division, 38. 6d. paper covers, or strongly bound in cloth, 4s. CASSELL'S LESSONS IN GERMAN. Parts I. and 11.-Price 2s. each in paper covers, or 28. 6d. in cloth. 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