Sept. 14, 1883.] 



• KNOWLEDGE ♦ 



175 



^ur iHatftematical Column* 



EASY LESSONS IN GEOMETKICAL PROBLEMS. 



By Richard A. Proctok. 



{Continued from page 127.) 



SUCH problems as I mentioned in my last, usually belong to a 

 more advanced stage of study than that for which these 

 simple papers are intended. They also often require the u.se of the 

 Si-xth Book. It will suffice here to consider a few of the simplest 

 cases. 



Suppose we have such a problem as this given : — 



Fig. 34. 



The sides of the trianrjle ABC (Fig. 34) are hisected in the points 

 a, b, c, and the three straight Ihies ak, b 1, and cm are draion at 

 right angles fo B C, A C, and A B respectively, shoiv that these three 

 straigh t lines, a k, b 1, and c m, pass through a point. 



Here the student might at once refer to the Fourth Book, and find 

 a proof in the cu'cumstance that a k and b I have there been shown 

 to meet at the centre of the circle through the points A, B, C. So 

 also by the same book do the liues a 1; and c m meet at the centre 

 of the circle through tlie points A, B, C. Now there is but one circle 

 passing through these points ; for if there were two, two circles 

 would intersect in three points, which is impossible. Hence a k, 

 b I, and c m pass through the same point. 



But although this proof is sound enough, it is not independent, 

 as a proof of this sort should be. Yet an actual and sufficient 

 ])roof will run closely, as might be expected, in the lines followed in 

 Book IV. 



It is hardly necessary to say that the proof must be indirect. 

 We can show, as in Book IV., that if ak and b I meet in 0, the 

 lines OA, OB, and OC are all equal. Then since AO = OB, a 

 line from O perpendicular to A B roiist bisect A B, in other words, 

 must pass through c, and coincide with c m. Hence if wo wished 

 to put the proof in Euclidean form, wc might begin by saying, — 

 If possible let cm not pass through the point O in which a k and 

 bl intersect, but have some other position as cmo. Then after 

 proving that A = O B, we could show that c is at right angles 

 to A B. But c iH is at riglit angles to A B, wherefore from the 

 same point c, there can be drawn two straight lines, at right angles 

 to A B and on the same side of it, — which is impossible, since all 

 right angles are ecjual. Therefore the line through c at right angles 

 to A B cannot lie otherwise than through 0. 



In a similar way we can deal with the problem — 



Ifilie three angles of the triangle ABC {Fig. 35), be bisected by the' 

 Hues A a, H >'. m,,! C •• these straight lines will all pass through one 

 point. 



But now suppose we have this problem, — 



From the angles A, B, and C, of the triangle, ABC, Fig 36, lines 

 are drawn at right angles fo B C, C A, amd AB respectively. These 

 three straight lines shall all pass through one point. 



Fig. 36. 



Here again the indirect method must be employed. We may 

 draw A a, B b, at right angles to B C, C A respectively, and inter- 

 secting in O ; then if we can prove that C (produced if necessary) 

 is at right angles to A B, what is required is done. 



We have in this case the angles at b and a right angles ; and it is 

 nearly always well to try in such cases whether any good comes 

 from noting that the angle in a semicircle is a right angle. This 

 at once shows that a circle on O C as diameter \n& pass through 

 bo; as will also a circle on A B as diameter. Suppose these 

 circles drawn ; or if any difiiculty arises from the effort to conceive 

 them as drawn, draw them in, as in the figure. 



Also it will obviously be convenient to draw in the lines ab,b c, 

 ca. 



We have now to show that C c is at right angles to A B. If 

 this be so, the angles c C A and c A C together make np a right 

 angle, or are complementary to each other. Of these the angle 

 c AC is a known angle; so that if we look for an angle known to be 

 complementary to c A C, we may be able to prove that so also is 

 c C A. Now the angle A B b is complementary to c A C by the 

 construction. Can we show that ZABb= ZcCA? We must try 

 our circles. We see that ZABb=ZboA on the same segment 

 A b ; and we see that Z b a A or b a O = Z b C on the same 

 segment b 0. This clearly serves our purpose. For we have 



ZbC0= ZboO= ZbBA = compt. of CAB 

 wherefore angle C c A is a right angle. 



{To be continued.) 



i&uv asaftist Column* 



By " Five op Clubs." 



WHIST QUESTIONS. 



Returning Partner's Suit. — " It hath been given out," " usually 

 return your partner's lead, unless you have a good suit of your 

 own." "what am I to understand by a "good suit" to return? 

 Suits that are considered good to load from a hand next to the 

 dealer are not always good suits to commence with, after having 

 won your partner's trick. Can suits be called good that require 

 more than two leads coming from a tlurd hand to establish ? 



MUKF. 

 [Cavendish says, retm-n only if you have a " very goi>d suit." 

 For my own part, I attach so much importance to kuow-ing early 

 where my partner's strength lies, that I should say to him, Show 

 your own suit if it is anything above medium strength, besides of 

 course being long. — FrvE ok CtUBS. 



Long Weak Suits. — Is there any use trying to establish a suit of 

 Five, headed by a Ten, with three or four small trumps in the same 

 hand? The greater number of gentlemen I have had for partners 

 during the ])ast twelve months have always led from their long suit, 

 and wluit is most extraordinary, none of them have played the game 

 less than " forty years." Muff. 



[Tlierc can be no doubt they are right in so leading. The object 



