180 



THE POPULAR EDUCATOR. 



self. 18. Loosing one's self. 19. Loose yourselves. 20. I may have 

 remained behind. 21. 1'ou loosed yourself. 22. He may have loosed 

 himself. 23. Let thorn both loose themselves. 24. You two might 

 loose yourselves. 25. Of one loosing himself. 26. You were loosing 

 yourselves. 27. Having loosed themselves. 28. They might loose 

 themselves. 29. We might have loosed ourselves. 



EXERCISE 77. ENGLISH-GREEK. 



1. Ai/otjunv. 2. AUOITO. 3. AVOIVTO. 4. fLUftrtiat. 5. Ai/oyuei/ot. 6. \vaaatic. 

 7. E\tvovro. 8. AurrjTcu. 9. Aureate. 10. \vaoiro. 11. Aue<70<a. 12. Av<rw- 

 /ue0a. 13. Avaovrai. 14. AVJJTUI. 15. AvaaiaOov. 16. Av<ry. 17. Ai/<7aff0a<. 



EXERCISE 78. GREEK-ENGLISH. 



1. He was rubbed. 2. Thou mayest be rubbed. 3. Thou wouldest 

 be rubbed. 4. He would be rubbed. 5. They two might have been 

 loosed. 6. They might have been loosed. 7. Let him be loosed. 8. 

 To have been loosed. 9. Being about to be loosed. 10. To have been 

 rubbed. 11. Being about to be rubbed. 12. Thou wast loosed. 13. 

 Ye were loosed. 14. Thou shalt be loosed. 15. We may have been 

 loosed. 16. We might have been loosed. 17. They may have been 

 loosed. 18. Having been loosed. 19. To be about to be loosed. 20. 

 Having been rubbed. 21. Let him be rubbed. 22. I have been loosed. 

 23. I had been loosed. 24. I shall have been loosed. 25. They have 

 been loosed. 26. They had been loosed. 27. Thou mightest have been 



EXERCISE 79. ENGLISH-GREEK. 



1. E\v9n- 2. A.v0rj. 3. AvOetn. 4. Tpi/3n<reTa<. 5. AK0)croi>Tai. 6. Erpi/Sr). 

 7. AcXujuui. 8. AeXu/aei/or, -i|r. 9. AeXi/o-oi/rui. 



EXERCISES IN EUCLID. III. 



PROPOSITION XIII. In a triangle ABC (Fig. 13), if B o, c o, 

 bisecting' the angle ABC, B c A, and meeting in o, be equal, then 

 shall A B be equal to A c. For since o B = o c, angle o B c = 

 angle o c B ; but angle o B c is equal to half-angle ABC, and 



angle o c B equals half-angle 

 A c B ; and, by Axiom 7, the 

 halves of equal things are equal ; 

 therefore angle A B c = angle 

 Acs. Therefore, by Euc. I. 6, 

 AB = AC. Q. E. D. 



PROPOSITION XIV. In a tri- 

 angle ABC, if BO, CO, bisect- 

 ing the angles ABC, BCA, and 

 meeting in o, bo equal, then A o 

 will bisect the angle B A c. 



Taking the figure of the last proposition, we have proved that 

 if OB, o c be equal, then A B, A c are equal. Hence, since 

 A B = A c, and A o is common, also base B o = base c o ; there- 

 fore, by Euc. I. 8, angle B A o = angle c A o ; -i.e., A o bisects 

 angle BAG. Q. E. D. 



PROPOSITION XV. In the figure of Euclid I. 1, if the circles 

 cut again in F (Fig. 14), and 

 c A produced meet the circle 

 again in H, then c H is greater 

 than c F. Join H F, A F ; then, 

 since A H is equal to A F, being 

 radii of the same circle (Def. 

 15), the angle AHF= angle 

 AFH (Euc. I. 5). But the 

 angle c F H is greater than 

 the angle AFH (Axiom 9) ; 

 therefore the angle CFH is 

 greater than the angle c H F. 

 Therefore, by Euc. I. 19, 

 the side c H is greater than the side c F. Q. E. D. 



Corollary. Hence it is obvious that the diameter is the 

 A longest line that can be drawn within a 



circle ; for if c A H be a diameter, and F 

 any point on the circumference, the angle 

 CFH will be greater than the angle AFH; 

 i.e., than the angle c H F. Therefore c H 

 will be greater than c F. 



PROPOSITION XVI. In the figure of 

 Euclid I. 5 (Fig. 15), prove that B G must 

 be greater than B c. By Euc. I. 17, any two 

 E angles of a triangle are together less than 

 two right angles ; therefore ABC and ACS 

 are together less than two right angles. But they are equal ; 

 therefore each of them is less than one right angle. Therefore 

 B c A is less than a right angle. 



Fig. 14. 



Fig. 15. 



Fig. 16. 



But by Euc. I. 13, angles BCA, BC o are together equal to 

 two right angles, and BCA has been proved less than a right 

 angle. Therefore B c G is greater than a right angle. 



Again, by Euc. I. 17, B c G and c G B are together less 

 than two right angles, and B c G is greater than one right angle ; 

 therefore B G c is less than a right angle. Therefore B c G is 

 greater than B G c ; therefore side B G is greater than B c. 

 Q. E. D. 



PROPOSITION XVII. In the figure of Euclid I. 16, if E c bo 

 equal to EF (Fig. 16), the angle A F 



ABC will be equal to the angle BCF. 

 For, by construction, A E = E c, 

 B E = E F ; therefore, if E c = E F, 

 A E, E c, B E, and E F are all equal 

 (Axiom 1). Also it is proved in 

 Euclid I. 16, that angle BAE=: 

 angle E c F. But since E A B is an 

 isosceles triangle, angle EAB = EBA 

 (Euc. I. 5) ; therefore the angle E c F 

 is equal to angle E B A. But be- 

 cause EEC is an isosceles triangle, 

 angle E c B angle EEC; and, from 

 above, angle E C F = angle E B A ; 

 therefore, adding equals to equals, 

 whole angle A B c = whole angle BCF (Axiom 2). 



PROPOSITION XVIII. In the figure of Euclid I. 22, if tht, 

 circles cut again in L (Fig. 17), D K shall be equal to D L. Join 

 F L, G L ; then, since F K, F L are radii of the same circle, F K = 

 F L (Def. 15) ; and, since G K, G L are radii of the same circle, 



GK = GL (Def. 15). 

 Hence, in the triangles 

 F G K, F G L, because 

 G K G L, and F G is 

 common, also base F K 

 = base F L, therefore 

 angle F G K angle 

 F G L (Euc. I. 8). 



Again, in the tri- 

 angles D G K, D G L, 

 Fig. 17. because G K = G L, and 



D G is common, also in- 

 cluded angle D G K has been proved equal to included angle 

 D G L, therefore, base D K = base D L. Q. E. D. 



NOTE ON EUCLID I. 22. Of the three straight lines, A, B, c, 

 it is necessary that any two should be greater than the third, 

 because, by Euc. I. 20, any two sides of a triangle are greater 

 than the third. The necessity will also appear from the figure ; 

 for if the large circle cut D E in M, G H must evidently be greater 

 than G M, or the small circle will not cut the large, and the con- 

 struction will fail. 



If G H be greater than G M, then G H is greater than the 

 difference of F M and F G ; c is greater than the difference of A 

 and B ; -i.e., C with either is greater than the third ; and since 

 A, B, c, are in order of magnitude, A with either of the others is 

 greater than the third. 



The proposition that any two sides of a triangle are greater 

 than the third is obvious if a straight line be denned as tho 

 shortest distance between two points ; from which it follows 

 that P, 1 Broken line between two points must be longer than 

 a str:,,: 4'iit line. 



Hence two sides c ^ A < 



are greater than 

 the third. 



PROPOSITION 

 XIX. At a 

 given point in 

 a given straight 

 line, to make an D 

 angle equal to a 

 given rectilineal 



This is the Fig. 18. 



same as Euc. I. 



23, but the solution there given being of little use for prac- 

 tical purposes, the following modified form of it is sug- 

 gested : Let A (Fig. 18) be the given point in the given 

 straight line A B, and D c E the given rectilineal angle. In c D, 

 c E take two points, D, E, cuch that c = c E, and join D K. 





