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 Bisection In a triangle (Posted on 2014-02-01)
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.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 Vector Solution Comment 1 of 1

the problem statement:

" |BC| does not equal |AB| ".

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

and

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

Equations (1) and (2) become

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

and

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.

QED

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

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