398 



THE POPULAR EDUCATOR. 



"What frantic fit, quoth he, has thus distraught 

 Thee, foolish mail, so rash a doom to give ? 

 What justice ever other judgment taught, 

 But he should die who merits not to live ? 

 None else to death this man despairing drive 

 But his own guilty mind, deserving death. 

 Is then unjust to each his due to give ? 

 Or let him die that loatheth living breath ? 

 Or let him die at ease that liveth here uneath ? 



" Who travels by the weary, wandering way, 

 To come unto his wished home in haste, 

 And meets a flood that doth his passage stay ; 

 Is not great grace to help him over past, 

 Or free bis feet that in the mire stick fast ? 

 Most envious man, that grieves at neighbours' good, 

 And fond that joyest in the woe thou hast, 

 Why wilt not let him pass, that long hath stood 

 TJpon the bank ? why wilt thyself not pass the flood ? 



*' He there does now enjoy eternal rest, 



And happy ease, which thou dost want and crave, 

 And further from it daily wanderest ; 

 What if some little pain tho passage have, 

 That makes frail flesh to fear the bitter wave ; 

 Is not short pain well borne that brings long ease ? 

 And lays the soul to sleep in quiet grave ? 

 Sleep after toil, port after stormy seas, 

 Ease after war, death after life, does greatly please." 



The Red Cross Knight is next led by Una to the House of 

 Holiness, which is described in an elaborate and beautiful 

 allegory, in which the contrast with the House of Pride is 

 forcibly brought out. Here the knight receives purification and 

 instruction. Thus fitted for his task, he encounters the great 

 dragon he had come to meet, and after a three days' combat 

 slays him. The book closes with the rejoicings over the 

 slaughter of the dragon and the release of his victims, and 

 with the marriage of the knight to Una. 



The outline which we have given of this book will enable the 

 student to form some idea of the character of Spenser's alle- 

 gory ; the detailed beauties of the poetry can be learned only 

 from the poem itself. 



EXERCISES IN EUCLID. VI. 



PROPOSITION XXXVII. If the diagonals of a four-sided 

 figure bisect each other, it is a parallelogram. 



Let AC, BD (Fig. 37), the diagonals of the four-sided 

 figure A B c D, intersect in o ; then if A o = o c, and B o = o D, 

 A D the figure is a parallelogram. For 



since A o = o c, and B o = o D, also 

 the opposite angles A o B, COD, are 

 equal (Euc. I. 15) ; therefore the tri- 

 angles are equal in every respect 

 (Euc. I. 4), and the angles are equal 

 which are opposite to the equal sides 

 in each i.e., the angle A B o is equal 

 to the angle o D c ; therefore, by 

 Euc. I. 27, A B is parallel to c D. By 



Fig. 37. 



an exactly similar course of reasoning, A r> may be proved parallel 

 to B c, that is, the figure A B c D is a parallelogram. Q. E. D. 



Corollary 1. If the diagonals be equal, as well as bisecting 

 each other, the figure is rectangular. For because the two tri- 

 angles ABC, DEC (Fig. 38) have two sides, A B, A c, equal to 

 DC, DB, each to each, and likewise A n 



their base B c common, therefore the 

 triangles are equal in every respect ; 

 therefore the angle ABC equals angle 

 B c D. But when a straight line falls 

 upon two parallel straight lines it makes 

 the two interior angles upon the same 

 side equal to two right angles (Euc. I. 

 29) ; therefore ABC, BCD are equal to 



two right angles ; and therefore, since they are equal, each is a 

 right angle, and the figure is rectangular. 



Corollary 2. Hence, obviously, as is proved otherwise in 

 Euc. III. 31, the angle in a semicircle is a right angle. 



PROPOSITION XXXVIII. If the diagonals of a four-sided 

 figure bisect each other at right angles, the figure is a rhombus. 



Let A c, B D (Fig. 39) the diagonals of a four-sided figure 

 & B c D, intersect at right angles in o, then if A o = o c, und 



B O = o D, the figure is a rhombus. As in the last proposition, 

 the figure is a parallelogram ; and therefore, the opposite sides, 

 A B, c D, and AD, B C, are equal ; and because A o equals o C, 

 and o B is common and at right angles, therefore base A B equals 

 base B c ; hence all the sides are equal, and the figure is a 

 rhombus. Q. E. D. 



Corollary. If the diagonals be also equal, it is a square; 

 for, as in Corollary 1 of the last propo- A D 



sition, the figure is rectangular, and a 

 rectangular rhombus is a square. 



PROPOSITION XXXIX. If a four- 

 sided figure have its opposite sides equal, 

 it is a parallelogram. 



Let the four-sided figure A B c D (Fig. 

 39) have its opposite sides equal, that 

 is, A B = c D, and A D = B c, then it 

 is a parallelogram. Join A c ; then 



B Fig. 39. 



since A B = c D, and B c = A D, also base A c is common to 

 the two triangles ABC, ADC, therefore, by Euc. I. 8, the tri- 

 angles are equal; hence the angle A c B equals angle CAD, there- 

 fore A D is parallel to B c (Euc. I. 27). But AD = BO by 

 construction ; therefore, by Euc. I. 33, C D is parallel to A B, 

 and A B c D is a parallelogram. Q. E. D. 



PROPOSITION XL. If A B, B c, c D, D A (Fig. 40), be the sides 



of a parallelogram taken in ' 

 order, and points E, F, G, H 

 be taken in them, such that 

 AE=CG, and BF = DH, 

 the figure E F G H shall be 

 a parallelogram. For since 

 the opposite sides of a 

 parallelogram are equal 

 (Euc. I. 34), A B = c D. 

 But by construction A E = 



Fig. 40. 



F 

 Fig. 41. 



CG, therefore remainder BE = DG (Axiom 3), and by con- 

 struction, B F = D H ; therefore in the two triangles, B E F, 

 D G H, the sides BE, B F in the one are equal to the sides 

 D G, G H in the other, and the included angles F B E, G D H are 

 equal, being opposite angles of a parallelogram ; therefore the 

 triangles are equal (Euc. I. 4), and base F E equals base G H. By 

 a similar course of reasoning G F = E H ; hence the four-sided 

 figure E F G H has its opposite sides equal, and is therefore a 

 parallelogram by the last proposition. A 



Q. E. D. 



PROPOSITION XLI. If ABC (Fig. 41) 

 be any triangle, and D E the line joining 

 the middle points of A B, A c, then shall 

 D E be parallel to B c, and B c shall be 

 double D E. 



Join BE, DC, and bisecting B c in F, 

 join D F, E F ; then, since the triangles 

 A D E, B D E are upon equal bases A D, 

 BD, and have a common vertex E, they are equal. Again, 

 because the triangles A D E, DEC are upon equal bases A E, 

 E c, and have a common vertex D, they are equal, hence A D B 

 is equal to both the triangles DEB and DEC;, ,hence, by 

 Axiom 1, triangle DEB equals triangle DEC, and they are 

 upon the same base D E ; hence they are between the same 

 parallels by proposition, therefore D E is parallel to B c. By 

 an exactly similar proof it may be shown that D F, E F are 

 parallel to A c, A B ; hence D F c E and E F B D are parallelo- 

 grams ; therefore, by Euc. I. 34, D E = F c, and D E = B F ; 

 hence D E equals half B c. Q. E. D. 



Corollary 1. It is obvious that the angles of the triangle 

 D F E are equal to the angles of the triangle BAG, for, by Euc. 

 I. 34, the angle F D E equals angle F c E, and angle FED equals 

 angle FED; therefore also, by Euc. I. 32, the third angle D F B 

 is equal to the third angle DAE. 



Corollary 2. It is obvious also that the sides of the triangle 

 D F E are each equal to half the corresponding sides of the 

 triangle B A c, and the area being equal to each of the triangles 

 B D F, F E c, A D E is equal to one-fourth of the whole triangle. 

 Hence, if with three straight lines, each of which is one-half the 

 corresponding side of a given triangle, another triangle be 

 formed, the area of the small triangle is one-fourth the area of 

 the given triangle. 



PROPOSITION XLII. If o (Fig. 42) be any point within a tri- 

 angle ABC, and D, E, the middle points of A B, A c, and F, G, the 



