Transitivity in Action

Transitivity in mathematics is a property of relationships in which objects of a similar nature may stand to each other. If whenever object A is related to B and object B is related to C, then the relation at hand is transitive provided object A is also related to C. Being a sibling is a transitive relationship, being a parent is not.

If a line l1 is perpendicular to another line l2 (the mathematical notation for which is l1l2) and, for a third line l3, we also have l2l3, then it is not true that l1l3. Thus the relationship of mutual orthogonality is not transitive. On the other hand, if a number A divides a number B (A|B) and B|C, then A|C. Thus the relation "is divisible by" is transitive.

Transitivity of one relation is so natural that Euclid stated it as the first of his Common Notions

Things which are equal to the same thing are also equal to one another.

In mathematical notations: if A = B and B = C, then necessarily A = C. The equality is a transitive relation! On the first glance this statement lacks content. If not yet reaching Descartes' sophistication, a fellow mutters "I am I" and then repeats this in wonder, the transitivity of equality will only imply exactly same banality: "I am I". No need to sound it the third time. So why Euclid and other mathematicians after him felt it necessary to explicitly state a seemingly vacuous property? The reason is of course that the same object may appear in different guises whose identity may not be either obvious or a priori known.

In passing, the veracity of the statement "I am I" is abstracted in mathematics into another property of equality: reflexivity, A = A. Not all relations are reflexive. For example, "is divisible by" is reflexive while "being perpendicular to" is not.

From Ceva's Theorem we know that some lines in a triangle meet at a point. Here I am going to establish those facts by a more conventional means - using the transitivity of equality. In the standard notations,

  • Angle bisectors ALa, BLb, and CLc divide respectively angles A, B, and C into two equal parts.
  • Perpendicular bisectors are erected at the midpoints Ma, Mb, and Mc perpendicular to the corresponding sides.
  • Altitudes AHa, BHb, CHc are perpendicular to the sides BC, AC, and AB, respectively, and pass through the opposite vertex of the triangle.
  • Medians AMa, BMb, CMc connect the vertices with midpoints of the opposing sides.

You may check out those lines and some others with an applet.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at, download and install Java VM and enjoy the applet.

Each family of lines consists of loci of points that satisfy certain conditions. The word locus (plural loci) in geometry substitutes for the word set used in other branches of mathematics.

The bisector ALa (or rather the whole line to which ALa belongs) is the locus of points equidistant from the two lines bb and cc defined by the sides AC and AB of ΔABC. In other words, ALa = {P: dist(P, bb) = dist(P, cc)}. The distance function dist(P, bb) here is the Hausdorff distance between the single point set {P} and the line (which is of course also a set of points) bb. The underlying distance between two points is Euclidean. This is the shortest distance from P to the line bb. Two lines ALa and BLb intersect at a point I. (I skip the proof that two angle bisectors can't be parallel.) For this point I, dist(I, bb) = dist(I, cc) but also dist(I, cc) = dist(I, aa). By the transitivity of equality, dist(I, bb) = dist(I, aa) which simply means that point I also lies on the third bisector CLc.

(More accurately, the locus of points equidistant from two intersecting lines consists of two perpendicular lines. In a triangle, we thus have 4 associated points: 1 incenter and 3 excenters. To get resricted to a single point - the incenter - we should have considered the angles of the triangle as defined by semi-infinite lines - rays.)

The perpendicular bisector, say pba, through the point Ma is the locus of points equidistant from points B and C: {P: dist(P, B) = dist(P, C)}. Similarly for the other two bisectors: pbb = {P: dist(P, C) = dist(P, A)} and pbc = {P: dist(P, B) = dist(P, A)}. If two bisectors pba and pbb intersect at a point O, then dist(O, B) = dist(O, C) but also dist(O, C) = dist(O, A) which, by transitivity, imply dist(O, B) = dist(O, A). Therefore, O also lies on the third bisector pbc.

Point O is equidistant from all three vertices. Thus it serves as the circumcenter of the triangle, the center of the circle (the circumcircle) that passes through all three vertices.

The case of altitudes reduces to the case of perpendicular bisectors with the following trick. Through each vertex draw a line parallel to the opposite side. These three lines form a triangle, A'B'C'. Altitudes of ΔABC serve as the perpendicular bisectors of ΔA'B'C'. Q.E.D.

To prove that the three medians intersect at a point, I refer to the notion of barycentric coordinates. For a given ΔABC, every point in the plane is associated with the unique triple (wA, wB, wC) with wA + wB + wC = 1. If all three numbers are positive, the point lies inside ΔABC. Now, AMa = {(wA, wB, wC): wB = wC} and BMb and CMc are defined similarly. It then follows that if G is the point of intersection of AMa and BMb, it also lies on CMc.

As an additional example, here is another proof of the fact that three chords formed by three intersecting circles meet at a point.

In the plane, a circle SR(C) with radius R and center C is defined as the locus of points located at distance R from C: S = SR(C) = {P: dist(P, C) = R}. For an arbitrary point P, let d denote the distance from P to C: d = dist(P, C). Then the expression d² - R² is known as the power of P with respect to the circle SR(C). If P has coordinates (x,y) and the center C coordinates (a,b) then the circle has the equation (x - a)² + (y - b)² = R², or SR(C) = {(x, y): (x - a)² + (y - b)² = R²}. The power of P with respect to the circle is then defined by the expression


Pow(P, S) = (x - a)² + (y - b)² - R²

Points with positive power lie outside the circle, those with negative power lie inside. The circle itself is the locus of points with zero power. Points that have the same power with respect to a circle lie on a concentric circle.

Let there be two nonconcentric circles (i.e., circles with different centers) S1 = SR1(C1) and S2 = SR2(C2). Then the locus of points that have the same power with respect to both circles is a straight line perpendicular to the center line C1C2. Using (1) we express Pow(P, S1) = Pow(P, S2) as


(x - a1)² + (y - b1)² - R1² = (x - a2)² + (y - b2)² - R2²

Obviously the square terms cancel out leaving a linear equation - an equation of a straight line. This straight line is called the radical axis of the two circles. The easiest way to see that the radical axis is perpendicular to the center line is to choose the coordinates so as to make the centers lie on the x-axis. Then b1 = b2 = 0 and, after cancelation, (2) does not contain y-terms which exactly means that the line is perpendicular to the x-axis.

Let now two circles S1 and S2 intersect at two points. Power of a point on a circle being 0, the two points of intersection of S1 and S2 obviously lie on the radical axis of the two circles. Therefore, the radical axis of two intersecting circles is the straight line that passes through their points of intersection. The problem of the three common chords then simply asserts that the pairwise radical axes of three intersecting circles meet at a point. This is the point that has the same power with respect to all three circles. From the foregoing discussion on transitivity, this is obvious, however, that, for any three circles not necessarily intersecting, the three radical axes meet at a point. The only restriction is that no two circles are concentric. The point is known as the radical center of the three circle. (There is a construction problem that is easily solved with the notion of radical center.)

(The idea of radical axis may also be introduced via the stereographic projection.)

Also, in any triangle, antiparallels to the sides adjacent to a vertex that cross on the symmedian through this vertex are equal. By transitivity then, the three antiparallels through the symmedian (Lemoine) point are equal.

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