itCISES IN EUCLID. 



US 



\omiiuitir* form. 



Singular. 

 >, /; 



til, t/iOU; 

 ( il. ht, it, iii. 



( elle, *hf, it, f. 



Plural, 

 noun, v* ; 

 TOUI. v< )U i V* ; 

 iln, m. they i 



ilar. 



me, me ; 

 te, 



f. they. 



(2.) Direct Object, Accusative. 

 When placed before the verb. 



Relative Form. 

 Singular. Plural, 



me, myMl/; nout, ourelM ; 



thyself; vou>, i/ourlri; 

 (him*. If; 



< he lie'/; M, thenueUw; 

 (iteelf; 



t--. 



*. 



toi, 



/If, him, if, m. ; 



j la, her, it, f. ; 



) s, himself, lierttlf. 



Plural, 

 noun, us; 



TOUB, IjOU ; 



lea, them ; 



so, tJiemseltes, one another, 



each other. 



When placed after the verb. 

 Singular. Plural, 



moi, me; nous, its; 



toi, (/.,,; TOUB, you; 



)e, him, it, m. ; 

 la, her, it, f. ; 



(3.) Indirect Object, Dative. 

 When placed before the verb. 



let, them, 



801, 



a soi, 



nx, m. ) 

 lles, f. J 



to himself ; { . j B e 



to herself; [ BUr 1 a e 

 to one's self; 

 to ittelf; j 

 (4.) Indirect Object., Genitive and Ablative. 

 Always placed after tho verb. 



to them. 



EXERCISES IN EUCLID. II. 



PROPOSITION VI. In the figure of Euc. I. 5, if o o (Fig. 6), 

 drawn at right angles to A G, meet A H in o, H being the inter- 

 section of B o and c F, then o F 

 ./ shall bo perpendicular to A F. 



|Jy In Proposition IV. wo proved 



c/'^V \ that A H will bisect the verti- 



.^y\ /*L-Jp"'*"'"" ca ^ an l B A c ; therefore the 

 ^^^^f\' angle a A o is = to angle O A p. 



^^ \J ^\\ Hence in the two triangles o AO, 



^^ -^ OAF, because a A = A F and 

 Fig. <j. A o is common, also included 



angle o A o = included angle 

 also base 00 = base o F (Euc. I. 4). Again, in tho 



OAT, 



same triangles, because 

 A o = A F, and o o = o F, 

 also base o A is common, 

 . . included angle AGO 

 '= included angle A F o. 

 But A G o is a right angle 

 by construction ; hence, 

 A F o is a right angle, 

 and o F is perpendicular 

 to A F. Q. E. D. 



. 



PBOPOSITJON VII. If A c (Fig. 7), the side of a triangle ABC, 



b bisected in D, and B D joined and produced to x, to that 

 D c may be equal to D B, then, if A B be joined, the angle A * i> 

 hall be equal to the angle D B c. 



Join c E. Then, since tho atraight lines A c, B E cat in i>, 

 therefore the angle A i> B in equal to the vertical and oppowiU 

 angle c D E (Euo. I. 1 5). Then in the two triangles A D B, c D z, 

 because side A D = D o by construction, and side B D = D K 

 also by construction, alao include- 1 , angle A o B = included 

 angle c D E, therefore base A B = c E. Again, in the aanx 

 triangles, because A B = c E, and B D = D E by construction, also 

 base A D = D c by construction, . . included angle A B D = 

 included angle DEC. By an exactly similar proof applied to tb 

 triangles ADZ, B D c, we see that 

 base A E = base B c, and angle A i> 

 = angle D B c. Q. E. D. 



PROPOSITION VIIL If A, B (Kg. 

 8) be two points on the same ride 

 of a given line c D, find in c D a 

 point E, such that the angle ABC 

 may be equal td the angle B B D. 



From A draw A c perpendicular 

 to c D (Euc. I. 12), and produce 

 A c to F, so that c F = A c (Post. 

 2, Euc. I. 3). Join B v (Post. 1), 

 Join AE (Post. 1), then shall the 

 For since A c ia at right 



Fig. 8. 



and let B F cut c D in E. 



angle A E c be = to angle BED. 



angles to E c, and by Euc. I. 15 the vertical or opposite angles 



between two straight lines are equal, therefore the angles 



ACE, E c F are equal, and A c = c F 



and c E is common, . . base A E = 



base E F (Euc. I. 4); and because 



A E = E F and E c is common, also 



base A c = base c F, . . angle A E c = 



angle c E F (Euc. I. 8). But by Euc. 



I. 15, angle c E F = angle BED; 



therefore, by Axiom 1, angle A E c 



angle BED. Q. E. F. 



PROPOSITION IX. In the figure of 



Euclid I. 1, if the circles cut again in F (Fig. 9), and A F, B r be 



joined, the figure A F B c is a rhombus. 



For since A c, A B, A F are radii of the Fame circle, A c and 



A F are equal to A B (Def. 15), and since B c, B A, B F are 



radii of the same circle, BC and B F are equal to B A (Def. 15). 



But by Axiom 1, things which are equal to the same thing ore 



equal to one another ; therefore A c, A F, B c, B F are all equal. 



Hence, by Def. 32, the figure A F B c is a rhombus. Q. E. 1). 

 PROPOSITION X. In an isosceles triangle A 



ABC (Fig. 10), if A L be drawn from the 



vertex A perpendicular to the bases c, and 



if A L be produced to M, so that L M = L A, 



then shall B M be = to B A. Join B M and 



M c. Then, because A L = L M by construc- 

 tion, and B L is common, also right angle 



B L A = right angle B L u, therefore base 



(Euc. I. 4). Q. E. D. 



PROPOSITION XI. If in any triangle the sides A B, A c (Fig. 



11) be bisected in L, u, and 

 L o, HO be drawn at right 

 angles to A B, A c, meeting in 

 o ; then o N drawn perpen- 

 dicular to B c will bisect B c. 

 Join o A, o B, O c. Thee, 

 because B L = L A, and L o 

 is common, also right angle 

 B L o = right angle A L o, 

 therefore base B O = base 

 o A (Euc. I. 4). In a simi- 

 lar way base c o = base o A 



(Euc. I. 4) ; therefore, by Axiom 1, B o = o c. Therefore, B o c 



is an isosceles triangle, and o N is drawn perpendicular to the 



base. Therefore, by Proposition I., o N bisects the base that 



isBN = NC. Q. E. D. 



Corollary. Hence, obviously o is tho centre of a circle passing 



through A, B, c. This is called the circumscribed circle, or circle 



described about the triangle ABC. 



PROPOSITION XII. In tho figure of Euclid I. 9, if with 



centre A (Fig. 12) and radius AF, a circle bo described cutting 



A B, A c in L and u, then shall E L be equal to D u. 



Fig. 10. 

 A = base B 



