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Bisection In a triangle (Posted on 2014-02-01) Difficulty: 2 of 5
The circle inscribed in the triangle ABC touches the side AC in the point K. Let the center of the circle be O, and the midpoint of AC be D. Prove that the line connecting O with D bisects the BK segment.

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Solution Vector Solution Comment 1 of 1

There is a caveat that should have been made about
the problem statement:

   " |BC| does not equal |AB| ".

A lemma about the incenter:

   AO = x*(AB/c + AC/b)


   AO = AB + y*(BA/c + BC/a)

   Since AB and AC are linearly independent, it is
   easy to solve for x and y to get

   AO  = (b*AB + c*AC)/(2*s)

   where s = (a + b + c)/2.

Now for the problem itself. Let E be the intersection
of lines BK and DO.

   AE = (1-x)*AB + x*AK                                         (1)


   AE = (1-y)*AD + y*AO                                         (2)

Equations (1) and (2) become

   AE = (1-x)*AB + x*([s-a]/b)*AC                           (1')


   AE = (1-y)*AC/2 + y*(b*AB + c*AC)/(2*s)           (2')

As in the lemma, it is easy to solve for x and y.
In that process we arrive at the following equation

   2*x*(c-a) = c-a

Here we use the caveat mentioned above and  are
able to cancel c-a from both sides and get

   x = 1/2

which in turn implies that E is the midpoint
of line segment BK.


  Posted by Bractals on 2014-02-02 17:09:06
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