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.
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)
and
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 = (1x)*AB + x*AK (1)
and
AE = (1y)*AD + y*AO (2)
Equations (1) and (2) become
AE = (1x)*AB + x*([sa]/b)*AC (1')
and
AE = (1y)*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*(ca) = ca
Here we use the caveat mentioned above and are
able to cancel ca from both sides and get
x = 1/2
which in turn implies that E is the midpoint
of line segment BK.
QED

Posted by Bractals
on 20140202 17:09:06 