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
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 AHa is the altitude from A to BC, then
Area(ABD) / Area(ACD) = [AHa·BD/2] / [AHa·CD/2] = BD / CD.
Combining the two gives the required identity:
AB / AC = BD / CD.
- 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
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.
18 January 2016, Created with GeoGebra
- 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 CLc, ∠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 ALa, BLb, and CLc 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: x12, x22, and x32. The midpoint of the arc x1x2 opposite the vertex x3 is then equal ±x1x2. And similarly for midpoints of arcs x2x3 and x1x3. Let's choose x1, x2, x3 such that all the signs are taken to be "+". The clinant of the line through x1x2 and x3 (the angle bisector at the vertex x3) is given by
M = (y32 - y1y2)/(x32 - x1x2) = - 1/(x1x2x32)
which readily gives an equation of the angle bisector
y = - (x - x1x2)/(x1x2x32) + y1y2
Write down a similar equation for a second angle bisector, for instance,
y = - (x - x1x3)/(x1x3x22) + y1y3
and solve the two equations for y. As x is the conjugate of y, with a little effort the result reduces to
x = - (x1x2 + x2x3 + x3x1)
a symmetric expression of all three indices. Therefore, this point also belongs to the third angle bisector.
- Liang-shin Hahn, Complex Numbers & Geometry, MAA, 1994
Copyright © 1996-2017 Alexander Bogomolny