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Inradii Ratio (Posted on 2014-03-08) Difficulty: 3 of 5

  
Let ABCD be a parallelogram. Let the incircle of ΔABC
touch diagonal AC at point P. Let r1 and r2 be the inradii
of triangles APD and PCD respectively.

            r1     |AP| 
Prove that ---- = ------
            r2     |PC|

  

See The Solution Submitted by Bractals    
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Solution Analytic Solution Comment 1 of 1
First the outline. 
Let A=(a,0) B=(0,b) C=(c,0) D=(a+c,-b) and c>a

P is a point on the x axis (p,0) where p is the weighted average the x coordinates of each of A, B, C by the lengths of the opposite sides.

AP/PC = (p-a)/(c-p)

The inradius of a circle is 2*Area/Perimeter
r1 = (p-a)*b/(BC+(p-a)+PD)
r2 = (c-p)*b/(AB +(c-p)+PD)

So for AP/PC = r1/r2 it suffices to show
BC + (p-a) = AB + (c-p)

Now for the messy algebra
AB=√(a+b)
BC=√(b+c)
p = [c√(a+b)+a√(b+c)]/[c-a+√(a+b)+√(b+c)]

The bold expressions above that must be shown to be equal both work out to
[a+b+c-ac+(c-a)(√(a+b)+√(b+c)]/[c-a+√(a+b)+√(b+c)]

Edited on March 11, 2014, 7:48 am
  Posted by Jer on 2014-03-10 23:45:18

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