June 29, 1883.] 



KNOWLEDGE ♦ 



397 



(Pur iflatbtmatical Column. 



GEOMETRICAL PROBLEMS. 

 By Richabd A. Proctor. 



WE will now try a problem of greater lengtli, thongh by no 

 means difBcnlt : — 



Ex. 5.— The triangles B A C, BBC (Fig. 10) arc on the same 

 hose, B C, and beliveen the same parallels, A D and B C ; also, B AC 

 is isosceles, the side B A being equal to the side A C. Show that 

 the perimeter of the triangle B A C is less than the perimeter of the 

 triangle B D C. 



r 



Fig. 10. 



First of all, we notice that B C being common to both triangles, 

 we need only prove that B A, A C are together less than B D, D C 

 together. 



After a little examination it becomes clear that the sides B A, A C 

 are not easily comparable with the sides B D, D C, as they stand. It 

 is an obvious resource to produce B A lo A F, making A P equal to 

 A C, so that B F is equal to the sura of the lines B A, A C ; and in 

 like manner to produce B D to G, making D G eqnal to D C, so that 

 B G is equal to the sum of the lines BD, D C. We have, then, to 

 show that B F is less than B G. If B F is less than B G, then 

 joining F G, the angle G is less than the angle B F G (Eu. I., 19). 

 But there seems no obvious method of proving this relation. 



Let ns consider our construction. We have A F equal to A C. 

 But A C is eqnal to A B. Thns B A, A C, and A F are all eqnal. 

 Hence a circle described with centre A, through B, would pass also 

 through C and F. Since B F would be the diameter of this circle, 

 B C F is a right angle. (Euc. 111., 31.) We therefore join C F, 

 and note that it is perpendicular to B C* 



We notice, also, that C A F is an isosceles triangle. Let us see, 

 then, about B G. We have D G eqnal to D C. But D C is less than 

 D B, since the angle D B C is less than the angle D C B. (Euc. I., 

 18.) This does not seem likely to help us. If we join G C, C D G 

 is, like C A F, an isosceles triangle. But this, again, does not 

 appear, on examination, to be a profitable relation. 



Let ns see, however, whether we are making use of all our data : 

 — We have been forgetting that A D is parallel to B C. Without 

 making nse of this relation, we cannot hope to solve our problem. 



We have shown that C F is at right angles to B C, and tlierefore, 

 of course, C F is at right angles to A D. It will be as well, there- 

 fore, to produce A D to meet F C in H, and to note tliat A H is at 

 right angles to F C. But C A F is an isosceles triangle, and it is a 

 well-known property that the line drawn from the vertex of a 

 right-angled triangle, at right angles to the base, bisects the base.f 

 Thus C H is equal to H F. Will this projierty help us ? Let ns 

 consider : C H is eqnal to H F, and H D A is at right angles to C F. 

 Clearly, then, if we join D F, we have D F equal to D C, for the 

 triangles D H C and D U F will be eqnal in all respects. 

 (Enc. I., 4.) 



* Hero is an instance of the advantage of carefully-constructed 

 figures. The relation arrived at by a tolerably obvions line of 

 reasoning might be overlooked for awhile. But if the figure has 

 been constructed carefully, in accordance with the data, the up- 

 rightness of F C could not escape notice, and a moment's inquiry 

 would show that it is not accidental, and suffice to exhibit its 

 cause. 



t This problem is not explicitly stated in Euclid. It is contained 

 implicitly in Bk. I., props. 10-12. It should be included amongst 

 the additional problems, a knowledge of which is necessary to those 

 who wish to work successfully at dedact'ions. 



D F being equal to D C, we may be led to proceed in one of two 

 ways : — 



First, we might notice that oar construction made D G equal to 

 D C ; so that D F is equal to D G, therefore the angle D F G equal to 

 the angle D F G (Euc. I., 5) ; therefore the angle BFG greater 

 than the angle G ; and B G greater than B F (Euc. I., 19) ; that is, 

 B 0, D 0, together greater than B A, A C together. 



Or, we might notice that since D F is equal to D C, B D and D P 

 arc together equal to B D and D C together ; but B D and D F are 

 together greater than B F (Euc. I., 2U) ; therefore B D and C D are 

 together greater than B A and A C together. 



If we had followed the first of these courses, wc should still 

 scarcely fail to notice a/tenoards that the second is an available 

 and a better solution. 



We proceed, then, to run over such steps of the above work as 

 are necessary to the proof of the proposition. In doing this we 

 notice that a property of the third book has been made use of in 

 proving that C F is at right angles to A H. We will assume that 

 the proposition has been given as a deduction from the first book. 

 Then, although the student might mentally have followed the course 

 we have adopted,* it would be well for him to modify the proof so 

 as to avoid the use of Book III. This is easily done. The student 

 sees at once that the proof involves the equality (in all respects) of 

 the triangles A H F, A H C. He had the angle A F U equal to 

 the angle A C II, A F equal to A 0, and A H common. 

 This is not quite sufficient (so far as Euclid's treatment of triangles 

 e.^itcnds). But it is easy to supplement these data by establishing 

 the equality of the angles FA II, H A C, these angles being respec- 

 tively equal to the equal angles A B C, A C B (Euc. I., 29). Hence 

 the triangles H A F, H A C are equal in all respects. 



The construction and proof of the proposition we are dealing^ 

 with run, therefore, thus : — 



Produce B A to F, making A F eqnal to A C. Join D F, D C, 

 and let A I), produced if necessary, meet P C in H. Then the 

 angle F A II is equal to the interior angle ABC (Enc. I., 29). But 

 A B C is equal to A C B (Euc. I., 5), and A C B to C A H (Euc. 1., 

 29). Therefore the angle F A H is equal to the angle C A H. Also, 

 F A, A H are equal to C A, A H respectively. Therefore the 

 triangles F A H, C A H are equal in all respects (Euc. I., 4). 

 Hence P H is equal to H C, and the angles at H are right angles. 

 Thus the triangles D H P and D H C are equal in all respects (Euc. 

 I., 4). Therefore D F is equal to D C. But B D and D F are 

 together greater than B F (Euc. I., 20), that is, than B A, A F 

 together. Therefore B D and D C are together greater than B A 

 and A C together, and the perimeter of B D C is greater than the 

 perimeter of B A C. 



(To he continued.) 



(Bur €l)tis Column, 



By Mephisto. 



AN INTEKSATIONAL TOURNAMENT IN GERMANY. 



WE have pleasure in announcing that the bi-annual International 

 Congress of the German Chess Association will take place at 

 Nuremberg, beginning on July 15. The prizes for the diief tourna- 

 ment are £00, £40, £25, £15, £10. Besides this tournament there 

 will be several minor tournaments, a problem tournament, solution 

 tournament, &c. It is expected that from this conntrj- several of 

 the strongest players will attend. Blackburno and Mason have 

 already determined upon doing so. It will be remembered that at 

 the last meeting at Berlin in 1881, Blackbume carried off first prize. 

 We hope to be specially represented at the tournament, and to give 

 our readers an ample report weekly. 



THE TOURNAMENT. 



Mackenzie won his game against Rosenthal on Wednesday 

 the 20th, as announced in our columns last week. This made 

 Blackburn absolute third prize winner. On Thursday English 

 was fortunate enongh to win from Blackbume, to whom the game 

 was of " no consequence," but it improved English's chance for a 

 prize; he stood at 14t. On Friday Rosenthal and English drew 



• More probably he would have noticed the equality of the 

 triangles A F H and A H C, if bis figure had been drawn with a 

 moderate amount of care. He would then see at a glance that this 

 equality was not accidental, would complete the proof mentally on 

 the assumption that H C is equal to H F; and in writing out the 

 proof he would give that proof of the equality of the triangles, 

 A n F and A H C, which follows above. 



