ADDITIONAL REMARKS ON CONCURRENCEIn an equilateral triangle, the centroid, circumcenter, orthocenter and incenter all coincide.The reason for this is simple: If we are given an equilateral triangle ∆ABC and the midpoints of[BC], [AC] and [AB] are given by D, E and F respectively, then(i) the ray [AD is also the angle bisector for6BAC, and the line AD is also the altitude fromA to BC and the perpendicular bisector of [BC] (all because d(A, B) = d(A, C)),(ii) similarly for the rays [BE and [CF as well as the lines BE and CF .However, in general the centroid, circumcenter, orthocenter and incenter of a triangle are all distinct,and in fact one can prove that if two of these points coincide then the triangle is equilateral. Theproof splits into six cases corresponding to the following hypotheses:(1) The centroid and orthocenter coincide.(2) The centroid and incenter coincide.(3) The centroid and circumcenter coincide.(4) The incenter and orthocenter coincide.(5) The circumcenter and incenter coincide.(6) The circumcenter and orthocenter coincide.PreliminariesBefore proving the theorem stated above, we shall establish some auxiliary results that weshall need.PROPOSITION. If L is a line, x is a positive real number, and X is a point not on L, thenthere are at most two points on L whose distance from X is equal to x.Proof. Suppose that B, C, D are three points on L such that d(X, B) = d(X, C) = d(X, D) = x.Relabeling the points if necessary, we may assume that B ∗ C ∗ D holds. Let E be the midpoint of[BC] and let F be the midpoint of [CD]. Since X is equidistant from B, C, D it follows that XEis the perpendicular bisector of [BC] and XF is the perpendicular bisector of [CD]. However, weknow that there is only one perpendicular from X to L, so this is a contradiction. The source ofthis contradiction is our assumption that there are three points on L which are equidistant fromX, and therefore we conclude that there are at most two such points.The second result analyzes pairs of triangles which satisfy SSA; as noted in Section II.4, thereis no general congruence theorem in this case, but the following result shows that there are at mosttwo possibilities:THEOREM. (SSA congruence ambiguity). (i) Suppose that we are given ∆ABC. Then thereis at most one point G ∈ (AC such that G 6= C and d(B, C) = d(B, G).(ii) If G is given as above and ∆DEF is a triangle such that d(A, B) = d(E, F ), d(B, C) =d(E, F ) and |6BAC| = |6EDF |, then either ∆ABC∼=∆DEF or else ∆ABG∼=∆DEF .Proof. (i) There is at most one other point G ∈ AC such that d(B, G) = d(B, C) by the precedingproposition, so there is at most one such point on (AC.1(ii) Let H ∈ (AC be such that d(A, H) = d(D, F ). Then by SAS we have ∆ABH∼=∆DEF ,and consequently we also have that d(B, H) = d(E, F ) = d(B, C). If H = C, then ∆ABC∼=∆DEF ; on the other hand, if H 6= C, then H must be the second point G which satisfies theconditions in (i), and we have ∆ABG∼=∆DEF .Remark. If BC ⊥ AC, then the only point G ∈ AC such that d(B, G) = d(B, C) is C itself,and this is why one has a hypotenuse-side congruence theorem for right triangles. On the otherhand, it is also possible that one has a second point G ∈ AC at the prescribed distance but Gdoes not lie on the open ray (AC; this happens if d(B, C) > d(A, B). On the other hand, if weare given ∆ABC such that d(A, B) > d(A, C) and AC is not perpendicular to BC, then there isa point G 6= C on (BC such that d(A, G) and d(A, C), which means that ∆ABC and ∆ABG arenot congruent even though they satisfy SSA.Proofs in the six individual casesIn all cases it will suffice to prove that d(A, B) = d(A, C), for if we know this we can alsoconclude that d(A, C) = d(C, A) = d(C, B) = d(B, C) by switching the roles of C and A in theappropriate discussion.One fact which is used repeatedly in our arguments is that the incenter and centroid of a trianglecan never lie on the triangle itself; in contrast, it is possible for the circumcenter or orthocenter tolie on the triangle, and in fact they always do so for right triangles.CASE (1): The centroid and orthocenter of ∆ABC coincide. Let D be the midpoint of[AC]. If the centroid and orthocenter are the same, then AD also contains the orthocenter; butthis means that AD ⊥ BC. Therefore we have d(B, D) = d(C, D), |6ADB| = 90◦= |6ADC|, andd(A, D) = d(A, D), so that ∆ADB∼=∆ADC by SAS. By the conclusion of the preceding sentenceit follows that d(A, B) = d(A, C).CASE (2): The centroid and incenter of ∆ABC coincide. We shall give two proofs; thefirst does not require the use of Playfair’s Postulate, but the second does.First proof. We shall suppose that d(A, B) 6= d(A, C) and derive a contradiction. Withoutloss of generality we may assume that d(A, B) < d(A, C) (we can dispose of the other case byswitching the roles of B and C in the argument that follows). Let D be the midpoint of [BC]; sincethe centroid and incenter coincide, we know that [BD is the angle bisector of6BAC.Notice that ∆ABD and ∆ADC satisfy the SSA conditions |6BAD| = |angleBAC|, d(A, D) =d(A, D) and d(B, D) = d(C, D). By itself this is not enough to prove that ∆ADB∼=∆ADC,which in our setting would be equivalent to showing that d(A, B) = d(A, C), so we really need todetermine whether d(A, B) < d(A, C) is possible. — Let E ∈ (AB be such that d(A, E) = d(A, C);the distance inequality implies that A ∗ B ∗ E must hold. We then have ∆DAC∼=∆DAE bySAS. This means that d(D, E) = d(D, C) = d(D, B), where the second inequality holds becauseD is the midpoint of [BC]. By the Isosceles Triangle Theorem it follows that |6DEB =6DEA|is equal to |6DBE =6CBE|, and by the previously established congruence relation we know that|6DEA| = |6ACB|. If we combine these, we see that |6EBC| = |6ACB|On the other hand, the Exterior Angle Theorem implies that |6EBC| > |6ACB|, so we havea contradiction. The source of this contradiction is our assumption that d(A, B) and d(A, C) areunequal, and therefore we must have d(A, B) = d(A, C).2Second proof. This uses the Angle Bisector Theorem from Section III.5 of the notes (whichdepends upon the theory of similar triangles and hence upon Playfair’s Postulate). — Let D be themidpoint of [AC]. Since the centroid and incenter are the same and neither lies on [AC], it followsthat [AD is the angle bisector for6BAC. Therefore the Angle Bisector Theorem implies thatd(A, C)d(C, B)=d(A, D)d(D, C)and since D is the