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 atangle𝐴𝐴𝐴𝐴! bisectors and we can be sure of this∘ because ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴∘ 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 , which implies that oncurrence of lines. ∡𝐼𝐼𝐼𝐼𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 �𝐼𝐼𝐼𝐼𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐼𝐼𝐴𝐴 . 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 Atriangle, 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. butangles 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: 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 A general remark.; see Figure 3. and contrast the approaches used to prove them. is the internal bisector of . strategiesPerpendicularor more used. lines. We bisectors study (The some dual problem: well known proving results the in collinearity this area 𝐏𝐏𝐏𝐏 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 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 ∡𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 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 ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 meet Perhapsand contrast the easiest the approaches of all results used on to concurrence prove them. is this: quantities 𝑢𝑢𝑢𝑢 and 𝑣𝑣𝑣𝑣 are equal, show that both of CperpendicularPerpendicular bisectors bisectors of the three sides of a triangleGiven concur. two at 𝐼𝐼𝐼𝐼. Draw perpendiculars 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 from 𝐼𝐼𝐼𝐼 to sides Perhaps the easiest of all results on concurrence is this: them are equal to a third quantity 𝑤𝑤𝑤𝑤. Viewed thus Cdistinct points and , the locus of points such thatGivenThe two 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 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: perpendiculardistinct 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 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 mathematics, at all levels. ThisPerhaps is best the proved easiest𝐀𝐀𝐀𝐀 using of𝐁𝐁𝐁𝐁 all resultsthe idea on of concurrence a locus, namely:𝐏𝐏𝐏𝐏 is this: and 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼. Hence 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼, which implies that 𝐼𝐼𝐼𝐼 Medians C is the perpendicular bisector of segment Given two 𝐏𝐏𝐏𝐏𝐀𝐀𝐀𝐀 𝐏𝐏𝐏𝐏𝐏𝐏𝐁𝐁𝐁𝐁 𝐀𝐀𝐀𝐀 𝐁𝐁𝐁𝐁 𝐏𝐏𝐏𝐏 𝐀𝐀𝐀𝐀𝐁𝐁𝐁𝐁; see lies on the internal bisector of ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. So the three distinct points and , the locus of points such that Figure𝐏𝐏𝐏𝐏𝐀𝐀𝐀𝐀 𝐏𝐏𝐏𝐏𝐏𝐏 1𝐁𝐁𝐁𝐁 (a). Here’s how the proof uses this locus idea.𝐀𝐀𝐀𝐀𝐁𝐁𝐁𝐁; 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 perpendicular𝐀𝐀𝐀𝐀 bisectors𝐁𝐁𝐁𝐁 of sides 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and𝐏𝐏𝐏𝐏 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 meet at 𝑂𝑂𝑂𝑂; see How can we be sure that the internal bisectors LetFigure𝐏𝐏𝐏𝐏𝐀𝐀𝐀𝐀 the𝐏𝐏𝐏𝐏𝐏𝐏 1 perpendicular𝐁𝐁𝐁𝐁 (b). Join 𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴𝑂𝑂𝑂𝑂𝑂𝑂 bisectors𝐴𝐴𝐴𝐴𝑂𝑂𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴. Then of sides𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝑂𝑂𝑂𝑂and𝐴𝐴𝐴𝐴, and𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 alsomeet𝐀𝐀𝐀𝐀𝐁𝐁𝐁𝐁𝑂𝑂𝑂𝑂;𝐴𝐴𝐴𝐴 at see𝐴𝐴𝑂𝑂𝑂𝑂;𝑂𝑂𝑂𝑂 see𝐴𝐴𝐴𝐴, of ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 do meet? The lines must A The medians of a triangle concur. henceFigure𝑂𝑂𝑂𝑂 1𝐴𝐴𝐴𝐴 (a).(b).𝐴𝐴 Here’s𝑂𝑂𝑂𝑂Join𝐴𝐴𝐴𝐴. The𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴 how𝑂𝑂𝑂𝑂𝑂𝑂 last𝐴𝐴𝐴𝐴 the𝑂𝑂𝑂𝑂𝑂𝑂 equality𝐴𝐴𝐴𝐴 proof. Then meansuses𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴𝐴𝐴 this that𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴 locus,𝑂𝑂𝑂𝑂 andis idea. equidistantalso 𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴𝐴𝐴𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴, from 𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴 and hence lies on the perpendicular bisector of 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. henceLet the𝑂𝑂𝑂𝑂 perpendicular𝐴𝐴𝐴𝐴𝐴𝐴𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴. The last bisectors equality of sidesmeans𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 thatand𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴is equidistantmeet at 𝑂𝑂𝑂𝑂; see Bisector of Therefore the three perpendicular bisectors meet in a point. ABC fromFigure𝐴𝐴𝐴𝐴 1and (b).𝐴𝐴𝐴𝐴 Joinand𝑂𝑂𝑂𝑂 hence𝐴𝐴𝐴𝐴𝑂𝑂𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴 lies𝑂𝑂𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴 on. Then the perpendicular𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴𝐴𝐴𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴, and bisectoralso 𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴𝐴𝐴 of𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴,. A Bisector of angle E HowhenceTherefore can𝑂𝑂𝑂𝑂𝐴𝐴𝐴𝐴 we𝐴𝐴 the be𝑂𝑂𝑂𝑂 three𝐴𝐴𝐴𝐴 sure. The perpendicular that last the equality perpendicular means bisectors that bisectors meet𝑂𝑂𝑂𝑂 is inequidistant a of point. sides 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 angle ACB Q F fromHowand 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 can𝐴𝐴𝐴𝐴 anddo we meet?𝐴𝐴𝐴𝐴 beand sure That’s hence that easy: lies the on perpendicular they the meet perpendicular because bisectors they bisector of are sides not of 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. I P andThereforeparallel𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 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 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴; and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 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 they𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 are andrespectively𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 do meet? perpendicular That’s easy: to they𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and meet𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 because; and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 theyand are𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 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 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴; and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 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 atangle𝐴𝐴𝐴𝐴! bisectors and we can be sure of this∘ because ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴∘ 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 , which implies that ∡𝐼𝐼𝐼𝐼𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 �𝐼𝐼𝐼𝐼𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐼𝐼𝐴𝐴 . 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 butangles 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 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀. A general remark. 𝐏𝐏𝐏𝐏 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 ∡𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 See Figure 2 (a). The above two proofs have a common theme; namely, to prove that two Let the internal bisectors of ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 meet quantities 𝑢𝑢𝑢𝑢 and 𝑣𝑣𝑣𝑣 are equal, show that both of at 𝐼𝐼𝐼𝐼. Draw perpendiculars 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 from 𝐼𝐼𝐼𝐼 to sides them are equal to a third quantity 𝑤𝑤𝑤𝑤. Viewed thus 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 respectively; see Figure 2 (b). Using in generality, we see a theme used frequently in the locus fact stated above, we see that 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 mathematics, at all levels. and 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼. Hence 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼, which implies that 𝐼𝐼𝐼𝐼 Medians lies on the internal bisector of ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. 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 ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and ∡𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 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

𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

ratio is repeated. If 𝑡𝑡𝑡𝑡 is very close to 𝑡𝑡𝑡𝑡, 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 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵 ○ × may want. Similarly, if 𝑡𝑡𝑡𝑡 is very close to 𝑡𝑡𝑡𝑡, the is a median, and 𝐺𝐺𝐺𝐺 is a point on 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵, ○ × B C ratio becomes very large, and it can get as large [𝐺𝐺𝐺𝐺𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴�𝐺𝐺𝐺𝐺𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡 𝐺𝐺𝐺𝐺, say. (ii) Since 𝐴𝐴𝐴𝐴𝐶𝐶𝐶𝐶 is a median, as we may want (there is no upper bound). and 𝐺𝐺𝐺𝐺 is a point on 𝐴𝐴𝐴𝐴𝐶𝐶𝐶𝐶, [𝐺𝐺𝐺𝐺𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡�𝐺𝐺𝐺𝐺𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡 𝐺𝐺𝐺𝐺, 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 𝑢𝑢𝑢𝑢 𝑡𝑡 𝑡𝑡𝑡𝑡�𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 and calculus class rather than a geometry class, we monotonic function 𝑣𝑣𝑣𝑣 𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡�, respectively. Since [𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋�𝐵𝐵𝐵𝐵𝑋𝑋, may say that the ratio 𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 is a strictly we have 𝑢𝑢𝑢𝑢𝑢𝑢𝑣𝑣𝑣𝑣 𝑢𝑢 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝑢𝑢 𝐺𝐺𝐺𝐺 𝑢𝑢 𝐺𝐺𝐺𝐺; hence: Proof based on areaof considerations.the position of 𝑡𝑡𝑡𝑡. This turns out to be more challenging to prove, Considering △𝐺𝐺𝐺𝐺�𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵 and △𝐺𝐺𝐺𝐺�𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 and invoking the 𝑢𝑢𝑢𝑢𝑢𝑢𝑣𝑣𝑣𝑣𝑣𝑣𝑢𝑢𝐺𝐺𝐺𝐺𝑢𝑢 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 [𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐶𝐶𝐶𝐶𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋𝑋�𝐶𝐶𝐶𝐶𝑋𝑋, we have unlike the perpendicular bisector of a line that 𝐺𝐺𝐺𝐺� divides segment 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 in the ratio 2∶1. of a triangle divides it into two triangles with equal uses one basic idea over and over again: 𝑢𝑢𝑢𝑢𝑢𝑢𝑣𝑣𝑣𝑣 𝑢𝑢 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝑢𝑢 𝐺𝐺𝐺𝐺 𝑢𝑢 𝐺𝐺𝐺𝐺; hence: segment or the bisector of an angle. We offer three area The So 𝐺𝐺𝐺𝐺� and 𝐺𝐺𝐺𝐺� divide 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 in the identical ratio proofs of concurrence. The �irst is based on a well 𝑢𝑢𝑢𝑢𝑢𝑢𝑣𝑣𝑣𝑣𝑣𝑣𝑢𝑢𝐺𝐺𝐺𝐺𝑢𝑢 segment joining the midpoints of two sides of a (2∶1). This means that they are the same point! . (See Figure 6. Observe the notation used known theorem of elementary geometry: triangle is parallel to and half the third side. In other words, the point where 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 and 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 meet carefully: if 𝑋𝑋𝑋𝑋 is any geometric �igure, then [𝑋𝑋𝑋𝑋𝑋𝑋 From the above equalities we deduce that midpoint theorem is identical to the point where 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 meet. denotes the area of 𝑋𝑋𝑋𝑋.) ThisRemark. obviously means that 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐴𝐴 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 concur. 𝐺𝐺𝐺𝐺 𝐺𝐺 𝐺𝐺𝐺𝐺𝐺𝐺 ThisProof is, based of course, on the the midpoint theorem.. Here’s how we invoke this idea. In Figure 7 we Note the reasoning involved. It is Now we argue algebraically. Let 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝐵𝐵 𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴. 𝑡𝑡𝑡𝑡𝑡𝑡 In have shown △𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴 with two medians 𝑡𝑡𝑡𝑡𝐵𝐵𝐵𝐵 and 𝐴𝐴𝐴𝐴𝐶𝐶𝐶𝐶 rather more subtle than the reasoning used in Since △𝐺𝐺𝐺𝐺𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 and △𝐺𝐺𝐺𝐺𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 have equal altitude, and Figure 4 (a) we have drawn two medians, 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 which intersect at 𝐺𝐺𝐺𝐺. The line through 𝑡𝑡𝑡𝑡 and 𝐺𝐺𝐺𝐺 is the proofs for concurrence of the perpendicular their bases are in the ratio 𝑡𝑡𝑡𝑡 𝐵𝐵 𝑡𝑡, their areas bear and 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶; they intersect at 𝐺𝐺𝐺𝐺�. Consider △𝐺𝐺𝐺𝐺�𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 then drawn; it intersects 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 at 𝐷𝐷𝐷𝐷. Note that there bisectors and the angle bisectors. The this same ratio to one another. Hence: and △𝐺𝐺𝐺𝐺�𝐵𝐵𝐵𝐵𝐶𝐶𝐶𝐶. Since 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐸𝐸 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 (midpoint theorem), is nothing being saidprove about 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 being a median underlying principle by which we concluded the two triangles are similar to each other, and (therefore) 𝐷𝐷𝐷𝐷 being the midpoint of 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡. 𝑢𝑢𝑢𝑢 𝑡𝑡𝑡𝑡𝑡𝑡 𝑢𝑢 that points 𝐺𝐺𝐺𝐺�𝐴𝐴 𝐺𝐺𝐺𝐺� are the same is illustrated in hence their sides are in proportion. Since Rather, we have to that 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 is a median. To � Figure 5. Let 𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵 be a given segment, and let 𝑃𝑃𝑃𝑃 be Similarly, the areas of △𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 and △𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷 bear this 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐸𝐸 � 𝐵𝐵𝐵𝐵𝐶𝐶𝐶𝐶 (midpoint theorem, again), it follows make sure we do not fall into the trap of assuming � a variable point located strictly in its interior. same ratio to one another. Hence: that 𝐺𝐺𝐺𝐺�𝐵𝐵𝐵𝐵 𝐸𝐸 � 𝐺𝐺𝐺𝐺�𝐵𝐵𝐵𝐵. Hence 𝐺𝐺𝐺𝐺� divides segment 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 Then, as implicitly that it is a median (and thus assuming (This means that 𝑃𝑃𝑃𝑃 cannot coincide with either the very thing we wish to prove), we have drawn in the ratio 2∶1. moves from towards , this ratio assumes 𝑢𝑢𝑢𝑢𝑢𝑢 𝑢𝑢 𝑢𝑢𝑢𝑢 𝑡𝑡 𝑡𝑡𝑡𝑡𝑦𝑦𝑢𝑢𝑢𝑢𝑢𝑢 𝑢𝑢 𝑢𝑢𝑢𝑢𝑦𝑦𝑢𝑢 every𝐴𝐴𝐴𝐴 or 𝐵𝐵𝐵𝐵 possible.) Consider positive the ratio value𝑡𝑡𝑡𝑡 exactly𝐸𝐸 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝑡𝑡𝑃𝑃𝑃𝑃 once.𝑃𝑃𝑃𝑃. it using a dashed line. In Figure 4 (b) we have drawn 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 and the 𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 By subtraction the above two equalities yield: remaining median 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴; theyA intersect at 𝐺𝐺𝐺𝐺�. A No The various lines drawn within △𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝐴𝐴𝐴𝐴 create six smaller triangles. Certain pairs ofA these are 𝑢𝑢𝐺𝐺𝐺𝐺 𝑡𝑡 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑦𝑦𝑦𝑦𝐺𝐺𝐺𝐺 𝑡𝑡𝐺𝐺𝐺𝐺 𝑡𝑡𝑡𝑡 𝐺𝐺 If AD is a median, then [ABD] =[ACD]. If K is F E E any point on AD, then [KBD] =[KCD] (since G 2 KD is a median of G1 K △KBC). Hence by subtraction, B C B D C [ABK] =[ACK ].

(a) (b) B = D = C

Figure 4. Concurrence of the medians of a triangle: proof using similar triangles Figure 6. Property of a median of a triangle

5352 At Right Angles | Vol. 3, No. 3, November 2014 Vol. 3, No. 3, November 2014 | At Right Angles 5352 Vol. 3, No. 3, November 2014 ∣ At Right Angles 3 4 At Right Angles ∣ Vol. 3, No. 3, November 2014 A P B

Figure 5. Variation of the ratio as moves along

𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

B = D = C

Figure 6. Property of a median of a triangle

53 At Right Angles | Vol. 3, No. 3, November 2014 Vol. 3, No. 3, November 2014 | At Right Angles 53 4 At Right Angles ∣ Vol. 3, No. 3, November 2014 A the medial lines of coincide with the medial lines of ⋆ It follows, by symmetry, that: lines of △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 is similar to 𝒯𝒯𝒯𝒯 but has

△𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. half� its� dimensions. This implies that We now make use of this observation in a [𝒯𝒯𝒯𝒯 ]= � [𝒯𝒯𝒯𝒯]. y x surprising way. ⋆ On the other hand, we just noted that the F E ⋆ Next, we note that △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 is similar to △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴: it medial lines of △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 coincide� with those of is a copy of △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 but with half its scale. It △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. This means that� 𝒯𝒯𝒯𝒯 coincides with 𝒯𝒯𝒯𝒯. y G x stands to reason, surely, that any genuine Therefore we have: [𝒯𝒯𝒯𝒯 ] = [𝒯𝒯𝒯𝒯]. geometric property exhibited by △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 will ⋆ How can we reconcile these two statements? also be exhibited by △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷. In particular we can u v There is clearly just one� way: it must be that be sure that if the medians of △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 concur, both [𝒯𝒯𝒯𝒯]=� 𝒯𝒯 and [𝒯𝒯𝒯𝒯 ]=𝒯𝒯. In other words, both B D C then so do the medians of △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷; and if the 𝒯𝒯𝒯𝒯 and 𝒯𝒯𝒯𝒯 have zero area. medians of △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 do not concur, then neither ⋆ But this is just another way of saying that the Figure 7. Making use of the area bisection property to prove concurrence do the medians of △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷. not Closingmedians remark.𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷 concur! ⋆ Let us now suppose now that the medians of The Mathematical Gazette △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 do meet in a point, and rather that One proposition, but three the situation is as depicted in Figure 8 (b). In entirely different proofs …. How would you But we have already shown that 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥. Hence that appeared in , in the that case, it must be that the medians contrast them? 𝑡𝑡𝑡𝑡 𝑥𝑥 𝑡𝑡. But this means that 𝐷𝐷𝐷𝐷 is the midpoint of 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵. November 2001 issue, written by Mowaffaq Hajja 𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷 enclose a triangle 𝒯𝒯𝒯𝒯 (shown with a Hence 𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷 is a median of △𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵. and Peter Walker, and titled “Why Must the In a future column, we will discuss an important heavy blue �illing). Triangle’s Medians Be Concurrent?” The central tool in the study of concurrence, which allows us It follows that the third median passes through idea is ingenious and subtle. ⋆ Since △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 is similar� to △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, we expect to deduce a vast number of concurrence results in the point of intersection of 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 and 𝐵𝐵𝐵𝐵𝐶𝐶𝐶𝐶. That is, the that the triangle 𝒯𝒯𝒯𝒯 created by the medial one stroke. Remark. medians of the triangle concur. Figure 8(a) shows a triangle 𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 and points Note the use of algebraic manipulations 𝐷𝐷𝐷𝐷𝐷𝐷 𝐵𝐵𝐵𝐵𝐷𝐷 𝐶𝐶𝐶𝐶 which are the midpoints of its sides 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵, in this proof. As such, this is not a “pure 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵, 𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵; the median 𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷 has been drawn, and the geometry” proof, and purists will frown at it. But medial triangle 𝐷𝐷𝐷𝐷𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 shown shaded. Here’s how the central idea is, surely, a pleasing one. We we argue. SHAILESH SHIRALI is Director of Sahyadri School (KFI), Pune, and Head of the Community Mathematics Centre in Rishi Valley School (AP). He has been closely involved with the Math Olympiad movement in India. He is hope you enjoyed this neat interplay of algebra ⋆ The midpoint theorem implies that �igure the author of many mathematics books for high school students, and serves as an editor for Resonance and andThird geometry. proof. At Right Angles. He may be contacted at [email protected]. 𝐷𝐷𝐷𝐷𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 is a parallelogram. Since the diagonals of We now offer a third proof which a parallelogram bisect each other, we deduce draws upon the midpoint theorem for its central that 𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷 bisects side 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 of △𝐷𝐷𝐷𝐷𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵. So the medial logic (just like the �irst proof shown above) but in line through 𝐴𝐴𝐴𝐴 of △𝐴𝐴𝐴𝐴𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 is the same line as the a very different way. It is taken from an article medial line through 𝐷𝐷𝐷𝐷 of △𝐷𝐷𝐷𝐷𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵. A A

F E F E

↗ T

B D C B D C

(a) (b)

Figure 8.

5554 At Right Angles | Vol. 3, No. 3, November 2014 Vol. 3, No. 3, November 2014 | At Right Angles 5554 Vol. 3, No. 3, November 2014 ∣ At Right Angles 5 6 At Right Angles ∣ Vol. 3, No. 3, November 2014 the medial lines of coincide with the medial lines of ⋆ It follows, by symmetry, that: lines of △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 is similar to 𝒯𝒯𝒯𝒯 but has

△𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. half� its� dimensions. This implies that We now make use of this observation in a [𝒯𝒯𝒯𝒯 ]= � [𝒯𝒯𝒯𝒯]. surprising way. ⋆ On the other hand, we just noted that the

⋆ Next, we note that △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 is similar to △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴: it medial lines of △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 coincide� with those of is a copy of △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 but with half its scale. It △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴. This means that� 𝒯𝒯𝒯𝒯 coincides with 𝒯𝒯𝒯𝒯. stands to reason, surely, that any genuine Therefore we have: [𝒯𝒯𝒯𝒯 ] = [𝒯𝒯𝒯𝒯]. geometric property exhibited by △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 will ⋆ How can we reconcile these two statements? also be exhibited by △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷. In particular we can There is clearly just one� way: it must be that be sure that if the medians of △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 concur, both [𝒯𝒯𝒯𝒯]=� 𝒯𝒯 and [𝒯𝒯𝒯𝒯 ]=𝒯𝒯. In other words, both then so do the medians of △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷; and if the 𝒯𝒯𝒯𝒯 and 𝒯𝒯𝒯𝒯 have zero area. medians of △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 do not concur, then neither ⋆ But this is just another way of saying that the do the medians of △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷. not Closingmedians remark.𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷 concur! ⋆ Let us now suppose now that the medians of △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 do meet in a point, and rather that One proposition, but three the situation is as depicted in Figure 8 (b). In entirely different proofs …. How would you that case, it must be that the medians contrast them? 𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷 enclose a triangle 𝒯𝒯𝒯𝒯 (shown with a In a future column, we will discuss an important heavy blue �illing). tool in the study of concurrence, which allows us

⋆ Since △𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 is similar� to △𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴, we expect to deduce a vast number of concurrence results in that the triangle 𝒯𝒯𝒯𝒯 created by the medial one stroke.

SHAILESH SHIRALI is Director of Sahyadri School (KFI), Pune, and Head of the Community Mathematics Centre in Rishi Valley School (AP). He has been closely involved with the Math Olympiad movement in India. He is the author of many mathematics books for high school students, and serves as an editor for Resonance and At Right Angles. He may be contacted at [email protected].

55 At Right Angles | Vol. 3, No. 3, November 2014 Vol. 3, No. 3, November 2014 | At Right Angles 55 6 At Right Angles ∣ Vol. 3, No. 3, November 2014