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Topic: to prove: rays bisecting 3 angles of a triangle meet at a single point
Replies: 6   Last Post: Feb 10, 1999 10:10 AM

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 AWatsonNY@aol.com Posts: 1 Registered: 12/6/04
Re: to prove: rays bisecting 3 angles of a triangle meet at a single point
Posted: Feb 10, 1999 10:10 AM

Eileen Stevenson wrote:
>
> I have a geometry puzzle that I cannot seem to prove.
> I believe that if you bisect each of the 3 angles of a triangle with 3
> rays, those rays will meet at a single point.
> But I can't prove it. Can anyone help?

This is called the Incenter, because it is the center of the circle that
can be drawn just tangent to the 3 sides of the triangle.

Let's call the triangle ABC, after its vertices. If we bisect angle ABC
(with vertex B), we'll have a ray that I will call ray BD, where D is
any random point on the ray. Note that if you drop a perpendicular from
D to side BA (extending side BA, if necessary) at a point, which we
could call E, and from D to side BC (again, if you need to, extend side
BC) intersecting side BC at point F, you will get two congruent
triangles BDE and BDF that are congruent because they have 2 congruent
acute angles (ABD and CBD), two congruent right angles (BDE and BDF) and
they have a hypotenuse in common (BD). Then, since those two triangles
are congruent, then DE and DF are congruent.

B E A
----------*------------------
\ * | /
\ * | D /
\ * /
\ / * /
\/ */
F \ /
\/
C
This proves that any point on an angle bisector is equidistant from the
two sides of the angle. The converse of this theorem can also be proved:
if point is equidistant from two sides of an angle, then it is on the
angle bisector of that angle. (Try it--it's not hard.)

Now if we make the angle bisector at vertex A, and call it ray AG, then
ray Ag and ray BD have to intersect somewhere. Let's call that point H.
H is now equidistant from sides BA and BC, since it is on angle bisector
BD. H is also equidistant from sides AB and AC, since it is on the angle
bisector of angle A.If we combine these two facts, we get that H is
equidistant from sides AC and BC. Thus, H is equidistant from the two
sides of angle ACB, therefore it is on the angle bisector of angle C.

I think that proves what you wanted.

> ( related: there are 2 other interesting points determined by a triangle:
> 1. The center: If you bisect a side of a triangle & draw a line
> passing through the bisecting point and the oposite corner of the triangle,
> then do the same with the other 2 sides, all three will intersect at a
> single point, the center of the triangle. This is generally NOT the same
>

This one is called the Centroid, or center of gravity of a triangle.
Those lines are called the medians.

> 2. The center of the circle defined by the triangle. 3 points
> determine a circle. If you bisect the 3 sides with perpendicular lines,
> they will meet at a point that determines the center of the unique circle
> specified by the triangle.
>

That point is called the circumcenter, because it is the point that is
the center of the circle that circumscribes (is drawn around) the
triangle.

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Date Subject Author
12/27/97 Eileen Stevenson
12/28/97 Michael Keyton
12/28/97 Floor van Lamoen
12/28/97 Guy F. Brandenburg
5/16/98 Imrich Miklosko
5/19/98 Guy F. Brandenburg
2/10/99 AWatsonNY@aol.com