# The Angle Bisectors

For every angle, there exists a line that divides the angle into two equal parts. This line is known as the *angle bisector*. In a triangle, there are three such lines. Three angle bisectors of a triangle meet at a point called the *incenter* of the triangle. There are several ways to see why this is so.

**Angle Bisectors as Cevians**

This is Corollary 2 of Ceva's theorem.

Let's here prove the required proportion. Let AD be the angle bisector of angle A. The area of a triangle can be computed in many ways. I'll use two of them to compute the areas of triangles ABD and ACD. Let a denote half the angle BAC. Then

Area(ABD) / Area(ACD) = [AB·AD·sin(a)/2] / [AC·AD·sin(a)/2] = AB / AC.

On the other hand, if AH

_{a}is the altitude from A to BC, thenArea(ABD) / Area(ACD) = [AH

_{a}·BD/2] / [AH_{a}·CD/2] = BD / CD.Combining the two gives the required identity:

AB / AC = BD / CD. (A dynamic illustration of a different proof can be found elsewhere. And there is another one.)

**Via Transitivity of Equality**

An angle bisector of an angle is known to be the locus of points equidistant from the two rays (half-lines) forming the angle. Existence of the incenter is then a consequence of the transitivity property of equality.

**Angle Bisectors as Axes of a 2-line**

If we adopt Frank Morley's outlook, transitivity of equality will still be present but only implicitly.

An angle bisector can be looked at as the locus of centers of circles that touch two rays emanating from the same point. In a triangle, there are three such pairs of rays. Pick any angle and consider its bisector. Circles that touch two sides of the angle have their centers on the bisector. Conversely, any point on the bisector serves as the center of a circle that touches both sides of the angle. Consider two bisectors of angles formed by the pair a and b and by the pair b and c. The circle with the center at the point of intersection of the two bisectors touches all three sides. In particular, it touches the sides a and c and, therefore, has its center on the bisector of the angle formed by these two sides.

**Angle Bisectors as Altitudes**

Altitudes of a triangle serve as angle bisectors of the associated orthic triangle. This association can be used in reverse.

Consider ΔABC. Through each vertex draw a line perpendicular to the corresponding angle bisector. These three lines will form a triangle - say ΔA'B'C'. Note that, since A'B' is perpendicular to CL

_{c}, ∠BCA' = ∠ACB'. The same is true of the pairs of angles at vertices A and B. Let's call this a*mirror*property. As we know, the orthic triangle of ΔA'B'C' has the mirror property. We'll make use of this observation shortly.What we have to show now is that the bisectors AL

_{a}, BL_{b}, and CL_{c}pass through the vertices A', B', and C', respectively. Assume to the contrary, that at least one of them does not pass through the corresponding vertex. Then the orthic triangle of ΔA'B'C' could not possibly coincide with ΔABC. But, assuming they are different, in the ΔA'B'C' there would be two distinct inscribed triangles (ABC and the orthic) that possess the mirror property. However, it can be shown that this is impossible. There is only one triangle with that property. (For other properties of the orthic triangle see the discussion on Fagnano's problem.)The proof is pretty simple.

It appears that angles in a triangle with the mirror property are not arbitrary. Count the angles in the diagram. (Two cases - of obtuse and acute-angle triangles - should be considered separately. In the former case, instead of talking of inscribed triangles, we should consider triangles with vertices on the sides, or the extensions thereof, of a given triangle.) Any triangle with the mirror property must have the same angles as the orthic triangle and have its sides parallel to the latter. As the third diagram suggests, this is clearly impossible for a triangle different from orthic. (Similar considerations worked in one of the proofs of Morley's Theorem.)

**Complex numbers**

As in the study of altitudes, let vertices of a given triangle be located on the unit circle. For convenience sake, lets take them to be squares of complex numbers: x

_{1}^{2}, x_{2}^{2}, and x_{3}^{2}. The midpoint of the arc x_{1}x_{2}opposite the vertex x_{3}is then equal ±x_{1}x_{2}. And similarly for midpoints of arcs x_{2}x_{3}and x_{1}x_{3}. Let's choose x_{1}, x_{2}, x_{3}such that all the signs are taken to be "+". The clinant of the line through x_{1}x_{2}and x_{3}(the angle bisector at the vertex x_{3}) is given byM = (y _{3}^{2}- y_{1}y_{2})/(x_{3}^{2}- x_{1}x_{2})= - 1/(x _{1}x_{2}x_{3}^{2})which readily gives an equation of the angle bisector

y = - (x - x

_{1}x_{2})/(x_{1}x_{2}x_{3}^{2}) + y_{1}y_{2}Write down a similar equation for a second angle bisector, for instance,

y = - (x - x

_{1}x_{3})/(x_{1}x_{3}x_{2}^{2}) + y_{1}y_{3}and solve the two equations for y. As x is the conjugate of y, with a little effort the result reduces to

x = - (x

_{1}x_{2}+ x_{2}x_{3}+ x_{3}x_{1})a symmetric expression of all three indices. Therefore, this point also belongs to the third angle bisector.

### References

- Liang-shin Hahn,
*Complex Numbers & Geometry*, MAA, 1994

|Contact| |Front page| |Contents| |Geometry| |Eye opener|

Copyright © 1996-2018 Alexander Bogomolny

64828963 |