# 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 *l _{1}* is perpendicular to another line

*l*(the mathematical notation for which is

_{2}*l*⊥

_{1}*l*)

_{2}*l*, we also have

_{3}*l*⊥

_{2}*l*,

_{3}*l*⊥

_{1}*l*.

_{3}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 *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*, *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*AL_{a}, BL_{b}, and CL_{c}divide respectively angles A, B, and C into two equal parts.*Perpendicular bisectors*are erected at the midpoints M_{a}, M_{b}, and M_{c}perpendicular to the corresponding sides.*Altitudes*AH_{a}, BH_{b}, CH_{c}are perpendicular to the sides BC, AC, and AB, respectively, and pass through the opposite vertex of the triangle.*Medians*AM_{a}, BM_{b}, CM_{c}connect the vertices with midpoints of the opposing sides.

You may check out those lines and some others with an 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 AL_{a} (or rather the whole line to which AL_{a} 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, _{a} = {P: dist(P, bb) = dist(P, cc)}._{a} and BL_{b} intersect at a point I. (I skip the proof that two angle bisectors can't be parallel.) For this point I, _{c}.

(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 pb_{a}, through the point M_{a} is the locus of points equidistant from points B and C: _{b} = {P: dist(P, C) = dist(P, A)}_{c} = {P: dist(P, B) = dist(P, A)}._{a} and pb_{b} intersect at a point O, then _{c}.

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 _{A}, w_{B}, w_{C})_{A} + w_{B} + w_{C} = 1._{a} = {(w_{A}, w_{B}, w_{C}): w_{B} = w_{C}}_{b} and CM_{c} are defined similarly. It then follows that if G is the point of intersection of AM_{a} and BM_{b}, it also lies on CM_{c}.

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 S_{R}(C) with radius R and center C is defined as the locus of points located at distance R from C: _{R}(C) = {P: dist(P, C) = R}.*power of P with respect to the circle* _{R}(C)._{R}(C) = {(x, y): (x - a)² + (y - b)² = R²}.

(1)

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) S_{1} = S_{R1}(C_{1}) and S_{2} = S_{R2}(C_{2}). Then the locus of points that have the same power with respect to both circles is a straight line perpendicular to the *center* line C_{1}C_{2}. Using (1) we express _{1}) = Pow(P, S_{2})

(2)

(x - a_{1})² + (y - b_{1})² - R_{1}² = (x - a_{2})² + (y - b_{2})² - R_{2}²

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. _{1} = b_{2} = 0

Let now two circles S_{1} and S_{2} intersect at two points. Power of a point
on a circle being 0, the two points of intersection of S_{1} and S_{2} 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.

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

65873135