Chapter 1. The Medial Triangle

Chapter 1. The Medial Triangle

2The triangle formed by joining the midpointsof the sides of a given triangle is called the medialtriangle. Let A 1 B 1 C 1 be the medial triangleof the triangle ABC in Figure 1. The sidesof A 1 B 1 C 1 are parallel to the sides of ABC and1half the lengths. So A 1 B 1 C 1 is the area of4ABC.In factarea(AC 1 B 1 ) = area(A 1 B 1 C 1 ) = area(C 1 BA 1 )Figure 1:= area(B 1 A 1 C) = 1 4 area(ABC).Figure 2:The quadrilaterals AC 1 A 1 B 1 and C 1 BA 1 B 1 are parallelograms. Thus theline segments AA 1 and C 1 B 1 bisect one another, and the line segments BB 1and CA 1 bisect one another. (Figure 2)Figure 3:Thus the medians of A 1 B 1 C 1 lie along the medians of ABC, so both trianglesA 1 B 1 C 1 and ABC have the same centroid G.

3Now draw the altitudes of A 1 B 1 C 1 from vertices A 1 and C 1 . (Figure3) These altitudes are perpendicular bisectors of the sides BC and ABof the triangle ABC so they intersect at O, the circumcentre of ABC.Thus the orthocentre of A 1 B 1 C 1 coincides with the circumcentre of ABC.Let H be the orthocentre of the triangle ABC,that is the point of intersection of the altitudesof ABC. Two of these altitudes AA 2 and BB 2are shown. (Figure 4) Since O is the orthocentreof A 1 B 1 C 1 and H is the orthocentre of ABCthen.|AH| = 2|A 1 O|The centroid G of ABC lies on AA 1 and|AG| = 2|GA 1 |Figure 4:. We also have AA 2 ‖OA 1 , since O is the orthocentre of A 1 B 1 C 1 . ThusHÂG = GÂ1O,and so triangles HAG and GA 1 O are similar.ThusSince HÂG = GÂ1O,|AH| = 2|A 1 O|,|AG| = 2|GA 1 |.AĜH = A1ĜO,i.e. H, G and O are collinear. Furthermore, |HG| = 2|GO|.ThusTheorem 1 The orthocentre, centroid and circumcentre of any triangleare collinear. The centroid divides the distance from the orthocentre tothe circumcentre in the ratio 2 : 1.

4The line on which these 3 points lie is called the Euler line of the triangle.We now investigate the circumcircle of the medial triangle A 1 B 1 C 1 . Firstwe adopt the notationC(ABC)to denote the circumcircle of the triangle ABC.Let AA 2 be the altitude of ABC from the vertex A.(Figure 5) ThenA 1 B 1 ‖AB and|A 1 B 1 | = 1 2 |AB|.In the triangle AA 2 B, AÂ2B = 90 ◦ and C 1 is the midpointof AB. Thus|A 2 C 1 | = 1 2 |AB|.Figure 5:Thus C 1 B 1 A 1 A 2 is an isoceles trapezoid and thus a cyclicquadrilateral. It follows that A 2 lies on the circumcircleof A 1 B 1 C 1 . Similarly for the points B 2 and C 2 whichare the feet of the altitudes from the vertices B and C.Thus we haveTheorem 2 The feet of the altitudes of a triangleABC lie on the circumcircle of the medial triangleA 1 B 1 C 1 .Let A 3 be the midpoint of the line segment AH joiningthe vertex A to the orthocentre H. (Figure 6) Thenwe claim that A 3 belongs to C(A 1 B 1 C 1 ), or equivalentlyA 1 B 1 A 3 C 1 is a cyclic quadrilateral.We have C 1 A 1 ‖AC and C 1 A 3 ‖BH, but BH⊥AC.Thus C 1 A 1 ⊥C 1 A 3 .Furthermore A 3 B 1 ‖HC and A 1 B 1 ‖AB.But HC‖AB, thus A 3 B 1 ⊥B 1 A 1 .Figure 6:

Thus, quadrilateral C 1 A 1 B 1 A 3 is cyclic, i.e. A 3 ∈ C(A 1 B 1 C 1 ).Similarly, if B 3 C 3 are the midpoints of the line segments HB and HC respectivelythen B 3 , C 3 ∈ C(A 1 B 1 C 1 ).Thus we haveTheorem 3 The 3 midpoints of the line segments joining the orthocentreof a triangle to its vertices all lie on the circumcircle of the medialtriangle.Thus we have the 9 points A 1 , B 1 , C 1 , A 2 , B 2 , C 2 and A 3 , B 3 , C 3 concyclic.This circle is the ninepoint circle of the triangle ABC.5Since C 1 A 1 B 1 A 3 is cyclic with A 3 C 1 ⊥C 1 A 1 and A 3 B 1 ⊥B 1 A 1 , then A 1 A 3is a diameter of the ninepoint circle. Thus the centre N of the ninepointcircle is the midpoint of the diameter A 1 A 3 . We will show in a little whilethat N also lies on the Euler line and that it is the midpoint of the linesegment HO joining the orthocentre H to the circumcentre O.Definition 1 A point A ′ is the symmetric point of a point A through a thirdpoint O if O is the midpoint of the line segment AA ′ . (Figure 7)Figure 7:We now prove a result about points lying on the circumcircle of a triangle.Let A ′ be the symmetric point of H through the point A 1 which is themidpoint of the side BC of a triangle ABC. Then we claim that A ′ belongsto C(ABC). To see this, proceed as follows.The point A 1 is the midpoint of the segmentsHA ′ and BC so HBA ′ C is a parallelogram. ThusA ′ C‖BH. But BH extended is perpendicular to ACFigure 8:

6and so A ′ C⊥AC. Similarly BA ′ ‖CH and CH is perpendicularto AB so BA ′ ⊥AB. ThusA ̂BA ′ = A ′ ĈA = 90 ◦ .Thus ABA ′ C is cyclic and furthermore AA ′ is a diameter.ThusA ′ ∈ C(ABC).Similarly the other symmetric points of H through B 1 and C 1 , which wedenote by B ′ and C ′ respectively, also lie on C(ABC).Now consider the triangle AHA ′ . The pointsA 3 , A 1 and O are the midpoints of the sidesAH, HA ′ and A ′ A respectively. Thus HA 1 OA 3 isa parallelogram so HO bisects A 3 A 1 . We sawearlier that the segment A 3 A 1 is a diameter ofC(A 1 B 1 C 1 ) so the midpoint of HO is also thenthe centre of C(A 1 B 1 C 1 ). Thus the centre N ofthe ninepoint circle, i.e. C(A 1 B 1 C 1 ) lies on theEuler line and is the midpoint of the segmentHO.Figure 9:Furthermore the radius of the ninepoint circle is one half of the radius Rof the circumcircle C(ABC).Now consider the symmetric points of H through the points A 2 , B 2 andC 2 where again these are the feet of the perpendiculars from the vertices. Weclaim these are also on C(ABC).Let BB 2 and CC 2 be altitudes as shownin Figure 10, H is the point of intersection.Then AC 2 HB 2 is cyclic so C 2 ÂB 2 + C 2 ĤB 2 =180 ◦ .Figure 10:

7But BHC = C 2 ĤB 2 soBHC = 180 ◦ − C 2 ÂB 2 which we write as= 180 ◦ − ÂBy construction, the triangles BHC and BA ′′ C are congruent, so BĤC =BA ′′ C. ThusBA ′′ C = 180 ◦ − Â,and so ABA ′′ C is cyclic. ThusSimilarly for B ′′ and C ′′ .A ′′ ∈ C(ABC)

8Returning to the symmetric points through A 1 , B 1 and C 1 of the orthocentre,we can supply another proof of Theorem 1.Theorem 1 The orthocentre H, centroid G and circumcentre O ofa triangle are collinear points.Proof In the triangle AHA ′ , the points Oand A 1 are midpoints of sides AA ′ and HA ′ respectively.(Figure 11) Then the line segments AA 1 andHO are medians, which intersect at the centroid G ′of △ AHA ′ and furthermore|G ′ H||G ′ O| = 2 = |G′ A||G ′ A 1 |But AA 1 is also a median of the triangle ABC so thecentroid G lies on AA 1 with|GA||GA 1 | = 2Figure 11:Thus G ′ coincides with G and so G lies on the line OH with|GH||GO| = 2Remark On the Euler line the points H (orthocentre), N (centre ofninepoint circle), G (centroid) and O (circumcircle) are located as follows:Figure 12:with |HN||NO| = 1. . .|HG|and|GO| = 2

9If ABCD is a cyclic quadrilateral, the four trianglesformed by selecting 3 of the vertices are called the diagonaltriangles. The centre of the circumcircle of ABCDis also the circumcentre of each of the diagonal triangles.We adopt the notational convention of denoting points associatedwith each of these triangles by using as subscript thevertex of the quadrilateral which is not a vertex of the diagonaltriangle. Thus H A , G A and I A will denote the orthocentre, thecentroid and the incentre respectively of the triangle BCD.(Figure 13)Figure 13:Our first result is about the quadrilateral formed by the four orthocentres.Theorem 4 Let ABCD be a cyclic quadrilateral and letH A , H B , H C and H D denote the orthocentres of the diagonaltriangles BCD, CDA, DAB and ABC respectively. Then thequadrilateralH A H B H C H Dis cyclic. It is also congruent to the quadrilateral ABCD.Proof Let M be the midpoint of CD and letA ′ and B ′ denote the symmetric points through Mof the orthocentres H A and H B respectively. (Figure14)Figure 14:The lines AA ′ and BB ′ are diagonals of the circumcircleof ABCD soABB ′ A ′ is a rectangle. Thus the sides AB and A ′ B ′ are parallel and of samelength.The lines H A A ′ and H B B ′ bisect one another so H A H B A ′ B ′ is a parallelogramso we get that H A H B is parallel to A ′ B ′ and are of the same length.ThusH A H B and AB are parallel and are of the same length.

10Similarly one shows that the remaining three sides of H A H B H C H O areparallel and of the same length of the remaining 3 sides of ABCD. Theresult then follows.The next result is useful in showing that in a cyclic quadrilateral, varioussets of 4 points associated with the diagonal triangles form cyclic quadrilaterals.Proposition 1 Let ABCD be a cyclicquadrilateral and let C 1 , C 2 , C 3 and C 4 be circlesthrough the pair of pointsA, B; B, C; C, D; D, Aand which intersect at points A 1 , B 1 , C 1 andD 1 . (Figure 15) Then A 1 B 1 C 1 D 1 is cyclic.Proof Let B 1 be the point where circlesthrough AB and BC meet. Similarlyfor the points A 1 , C 1 and D 1 . Join thepoints AA 1 , BB 1 , CC 1 and DD 1 , extendto points Ã, ˜B, ˜C and ˜D. (These extensionsare for convenience of referring to angleslater.)Figure 15:We will apply to the previous diagram theresult that if we have a cyclic quadrilateral thenan exterior angle is equal to the opposite angleof the quadrilateral. In Figure 16, if CB is extended to ˜B then:˜B ̂BA + A ̂BC = 180 ◦C ̂DA + A ̂BC = 180 ◦Thus ˜B ̂BC = C ˜DA. . .In Figure 17 we apply the above result to 4 quadrilaterals.

11Figure 16:In C 1 CDD 1˜CĈ 1 D 1 = C ̂DD 1˜D ̂D 1 C 1 = C 1 ĈDIn BCC 1 B 1˜CĈ 1 B 1 = C ̂BB 1˜D̂B 1 C 1 = C 1 ĈBIn BB 1 A 1 A˜B ̂B 1 A 1 = BÂA 1In AA 1 D 1 DÃA 1 B 1 = A ̂BB 1Adding up anglesÃA 1 D 1 = A ̂DD 1˜DD 1 A 1 = DÂA 1Figure 17:C 1 ̂D1 A 1 + C 1 ̂B1 A 1 = C 1 ̂D1 ˜D + ˜D ̂D1 A 1 + C 1̂B1 ˜B + ˜B̂B1 A 1= C 1 ĈD + A 1 ÂD + C 1 ĈB + BÂA 1SimilarlyThus A 1 B 1 C 1 D 1 is cyclic.= C 1 ĈD + C 1 ĈB + A 1 ÂD + BÂA 1= BĈD + BÂD + 180◦D 1 Ĉ 1 B 1 + D 1 Â 1 B 1 = 180 ◦□

12We now apply this to orthocentres and incentresof the diagonal triangles.Theorem 5 Let ABCD be a cyclic quadrilateral and let H A , H B , H Cand H D be the orthocentres of the diagonal triangles BCD, CDA, DAB andABC, respectively. Then H A H B H C H D is a cyclic quadrilateral.Figure 18:Proof In the triangle BCD, (Figure 18),CĤAD = 180 ◦ − C ̂BD.In the triangle CDA, CĤBD = 180 ◦ − D ̂BCButDÂCCĤAD= D ̂BC so= CĤBDand so we conclude that CDH B H A is cyclic. Similarly, show that H B DAH C , H C ABH Dand H D BCH A are cyclic.Now apply the proposition to see that H A H B H C H D is cyclic.Theorem 6 If ABCD is a cyclic quadrilateral andif I A , I B , I C , I D are the incentres of the diagonal trianglesBCD, CDA, DAB and ABC, respectively, then the pointsI A , I B , I C and I D form a cyclic quadrilateral.Figure 19:

13Proof Recall that if ABC is a triangle, I is the incentreand P, Q, R are feet of perpendiculars from I to thesides BC, CA and AB thenBÎC= 1 (RIP + P IQ)2= 1 2 (360◦ − RIQ)= 1 2 (360◦ − 180 ◦ + RÂQ)= 90 ◦ + RÂQ .2Now apply this to the triangles BCD and ACD. We get(Figure 20) :Figure 20:CÎADCÎBD= 90 ◦ + 1 (C ̂BD),2= 90 ◦ + 1 2 (CÂD).But C ̂BD = CÂD so it follows that CÎAD = CÎBD. Thus I A CDI B is acyclic quadrilateral. The proof is now completed as in previous theorem.Theorem 7 If ABCD is a cyclic quadrilateral and G A , G B , G C andG D are the centroids of the diagonal triangles BCD, CDA, DAB and ABC,respectively, then the quadrilateral G A G B G C G D is similar to ABCD. Furthermore,the ratio of the lengths of their corresponding circles is 1 3 .

14Proof Recall that the quadrilateral formed by the orthocentres H A H B H C H Dis congruent to ABCD.We also have the fact that all four diagonal triangleshave a common circumcentre which is the centreof the circle ABCD. Let this be denoted byO.Now join O to H A , H B , H C and H D . (Figure 21) Thecentroids G A , G B , G C and G D lie on these line segmentsand|OG A ||OH A | = |OG B||OH B | = |OG C||OH C | = |OG D||OH D | = 1 3Figure 21:Then it follows that G A G B G C G D is similar to H A H B H C H D .The result now follows.Finally we haveTheorem 8 Let ABCD be a cyclic quadrilateral and let A 1 and C 1be the feet of the perpendiculars from A and C, respectively, to the diagonalBD and let B 1 and D 1 be the feet of the projections from B and D onto thediagonal AC. Then A 1 B 1 C 1 D 1 is cyclic.ProofConsider the quadrilateral BCB 1 C 1 (Figure 22). SinceFigure 22:

15BĈ1C = B ̂B 1 C = 90 ◦ , then BCB 1 C 1 is cyclic.Similarly, A, A 1 , B 1 , B are cyclic, C, C 1 , D 1 , D are cyclic and A, A 1 , D, D 1are cyclic. Then by results of Proposition 1, A 1 B 1 C 1 D 1 is cyclic.

16Four Circles TheoremIf 4 circles are pairwise externally tangent, then the pointsof tangency form a cyclic quadrilateral.In Figure 23, the quadrilateral ABCD is cyclic.Remark A similar theorem could not be true for 5 circlesas 3 of the intersection points may lie on a line.Figure 23:Figure 24: ABC lie along a lineProof Recall that if T K is tangent to circle atT and O is centre of circle, then the angle between chordT L and tangent line T is one half of angle subtended atcentre O by chord T L, i.e. KT L = 1 (T ÔL). This2is because K ̂T L = T ̂RL where T R is the diameter atT .Draw tangent lines AA 1 , BB 1 , CC 1 and DD 1 at points of contactwith A 1 , B 1 , C 1 , D 1 being points in region bounded by the circles.ThusBÂD + BĈD= BÂA 1 + A 1 ÂD + BĈC 1 + C 1 ĈD= 1 2 (AÔ1B + AÔ2D + BÔ3C + CÔ4D)= 1 2 (sum of angles of quadrilateral O 1O 2 O 3 O 4 )= 180 ◦ .Figure 25:

17Here we used the fact that the point of tangency of a pair of circleslies on the line joining their centres.RemarkthenIf r 1 , r 2 , r 3 and r 4 denote the radii of the four circlesO 1 O 2 + O 3 O 4 = r 1 + r 2 + r 3 + r 4and O 1 O 3 + O 3 O 4 = r 1 + r 2 + r 3 + r 4Thus O 1 O 2 O 3 O 4 is a quadrilateral with the sums of the oppositeside lengths equal. Such a quadrilateral is called inscribable,i.e. it has an incircle. In this situation, the circumcircle of ABCDis the incircle of O 1 O 2 O 3 O 4 .