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How to Prove It A ←Locus P Perp bisector How To of side AC Perp bisector How To of side AB − − ProveHow To It O A = = B B C ThisProve continues the ‘Proof’ It column begun earlier. In this (a) (b) This‘episode’ continues we study the some ‘Proof’ results column from begun geometry earlier. related In this to Figure 1. Concurrence of the perpendicular bisectors of the sides of a triangle ‘episode’the theme we of study concurrence. some results from geometry related to theThis theme continues of concurrence. the ‘Proof’ column begun earlier. In this ‘episode’ we study some results from geometry related to the theme of concurrence. certainly not parallel to each other — after all, meet because they are not parallel to each other, in the classroom Internalthey meet atangle! bisectors and we can be sure of this∘ because ∡ ∡∘ , which implies that oncurrence of lines. ∡ ∡ . oncurrence of lines. The internal bisectors of the three angles An extremely common theme in Given any ofOf a the triangle same nature, concur. and proved the same way, is Shailesh A Shirali triangle,A small tweak the external in this argument bisectors yields of any a relatedtwo of its plane geometry is thatAn of proving extremely the common concurrence theme of inthree this: oncurrence of lines. anglesbut slightly and less the internalfamiliar result: bisector of the third planeor more geometry lines. (The is that dual of problem: proving the proving concurrence the collinearity of three of three or more points.) It is of interest to study the different Given an angle , the locus of angle concur or more lines. (The dualAn problem: extremely proving common the themecollinearity in To show this, we use a different locus fact which strategies used. We study some well known results in this area points equidistant from the arms and of threeplane or more geometry points.) is thatIt is of interestproving tothe study concurrence the different of three just as basic: A general remark.; see Figure 3. and contrast the approaches used to prove them. is the internal bisector of . Perpendicularstrategiesor more used. lines. We bisectors study (The some dual problem: well known proving results the in collinearity this area and contrast the approaches used to prove them. The above two proofs have a Perpendicularof three or more points.) bisectors It is of interest to study the differentThe ∡ See Figure 2 (a). common theme; namely, to prove that two perpendicularstrategies used. bisectors We study of some the well three known sides ofresults a triangle in this concur. area The Let the internal bisectors of ∡ and ∡ meet andPerhaps contrast the easiest the approaches of all results used on to concurrence prove them. is this: quantities and are equal, show that both of CPerpendicularperpendicular bisectors bisectors of the three sides of a triangleGiven concur. two at . Draw perpendiculars from to sides Perhaps the easiest of all results on concurrence is this: them are equal to a third quantity . Viewed thus Cdistinct points and , the locus of points such thatGivenThe two , , respectively; see Figure 2 (b). Using in generality, we see a theme used frequently in This is best proved using the idea of a locus, namely: distinctperpendicular pointsis the bisectors perpendicularand , the of the locus bisector three of points sides of segment of asuch triangle that concur. the locus fact stated above, we see that mathematics, at all levels. ThisPerhaps is best the proved easiest using of all resultsthe idea on of concurrence a locus, namely: is this: and . Hence , which implies that Medians C is the perpendicular bisector of segment Given two ; see lies on the internal bisector of ∡. So the three distinct points and , the locus of points such that Figure 1 (a). Here’s how the proof uses this locus idea.; see internal angle bisectors meet in a point. This is bestisproved the perpendicular using the idea bisector of a locus, of segment namely: Figure 1 (a). Here’s how the proof uses this locus idea. TheTheorem. result generally encountered next is: Let the perpendicular bisectors of sides and meet at ; see How can we be sure that the internal bisectors LetFigure the 1 perpendicular (b). Join bisectors. Then of sidesand, and alsomeet; at see; see, of ∡ and ∡ do meet? The lines must A The medians of a triangle concur. henceFigure 1 (b).(a). Here’sJoin. The how last the equality proof. Then meansuses this that locus, andis idea. equidistantalso , from and and hence lies on the perpendicular bisector of . henceLet the perpendicular. The last bisectors equality of sidesmeans thatandis equidistantmeet at ; see Bisector of Therefore the three perpendicular bisectors meet in a point. ABC fromFigure 1and (b). Joinand hence lies on. Then the perpendicular, and bisectoralso of,. A Bisector of angle E HowhenceTherefore can we the be three sure. The perpendicular that last the equality perpendicular means bisectors that bisectors meet is inequidistant a of point. sides angle ACB Q F Howandfrom can anddo we meet? beand sure That’s hence that easy: lies the on perpendicular they the meet perpendicular because bisectors they bisector of are sides not of . I P andparallelTherefore do to theeach meet? three other, That’s perpendicular and easy: this they is ensured meetbisectors because by the meet fact they in that a are point. they not are respectively perpendicular to and ; and and are HowparallelKeywords: can to weConcurrence, each be sureother, that locus, and the this ratio, perpendicular is intersection, ensured by area, bisectors the monotone, fact that of sidesmonotonic they are andrespectively do meet? perpendicular That’s easy: to they and meet because; and theyand are notare Keywords: Concurrence, locus, ratio, intersection, area, monotone, monotonic BCR B D C parallel to each other, and this is ensured by the fact that they are respectively perpendicular to and ; and and are Keywords: Concurrence, locus, ratio, intersection, area, monotone, monotonic (a) (b) Figure 2. Concurrence of the internal bisectors of the angles of a triangle 5150 At Right Angles | Vol. 3, No. 3, November 2014 Vol. 3, No. 3, November 2014 | At Right Angles 5150 2 At Right Angles ∣ Vol. 3, No. 3, November 2014 A ←Locus P Perp bisector AC Perp bisector of side of side AB − − O A = = B B C (a) (b) Figure 1. Concurrence of the perpendicular bisectors of the sides of a triangle certainly not parallel to each other — after all, meet because they are not parallel to each other, Internalthey meet atangle! bisectors and we can be sure of this∘ because ∡ ∡∘ , which implies that ∡ ∡ . The internal bisectors of the three angles Of the same nature, and proved the same way, is Given any of a triangle concur. A small tweak in this argument yields a related this: triangle, the external bisectors of any two of its anglesbut slightly and less the internalfamiliar result: bisector of the third Given an angle , the locus of angle concur pointsTo show this,equidistant we use a from different the arms locus factand which ; see Figure 3. isjust the as internal basic: bisector of . A general remark. ∡ See Figure 2 (a). The above two proofs have a common theme; namely, to prove that two Let the internal bisectors of ∡ and ∡ meet quantities and are equal, show that both of at . Draw perpendiculars from to sides them are equal to a third quantity . Viewed thus , , respectively; see Figure 2 (b). Using in generality, we see a theme used frequently in the locus fact stated above, we see that mathematics, at all levels. and . Hence , which implies that Medians lies on the internal bisector of ∡. So the three internal angle bisectors meet in a point. TheTheorem. result generally encountered next is: How can we be sure that the internal bisectors of ∡ and ∡ do meet? The lines must A The medians of a triangle concur. Bisector of ABC A Bisector of angle E angle ACB Q F I P BCR B D C (a) (b) Figure 2. Concurrence of the internal bisectors of the angles of a triangle 51 At Right Angles | Vol. 3, No. 3, November 2014 Vol. 3, No. 3, November 2014 | At Right Angles 51 2 At Right Angles ∣ Vol. 3, No. 3, November 2014 I B A P B A ● ● Figure 5. Variation of the ratio as moves along ratio is repeated. If is very close to , the ratio immediately seen to have equal area, by making is close to 0, and it can get as close to 0 as we use of the property mentioned above. (i) Since ○ × may want. Similarly, if is very close to , the is a median, and is a point on , ○ × B C ratio becomes very large, and it can get as large [� , say. (ii) Since is a median, as we may want (there is no upper bound). and is a point on , [[ , say. In Figure 7 we have written these symbols within Figure 3. Concurrence of two external angle bisectors and one internal bisector Or, to use terminology generally heard in a the respective regions. Let [ and calculus class rather than a geometry class, we monotonic function �, respectively. Since [�, may say that the ratio is a strictly we have ; hence: Proof based on areaof considerations.the position of . This turns out to be more challenging to prove, Considering △� and △� and invoking the We offer a because the median is not so obviously a locus, midpoint theorem twice, like earlier, we deduce A median second proof which is of a very different nature. It Again, since [�, we have unlike the perpendicular bisector of a line that � divides segment in the ratio 2∶1. of a triangle divides it into two triangles with equal uses one basic idea over and over again: ; hence: segment or the bisector of an angle.
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