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|

  

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=√(a²+b²)
BC=√(b²+c²)
p = [c√(a²+b²)+a√(b²+c²)]/[c-a+√(a²+b²)+√(b²+c²)]

The bold expressions above that must be shown to be equal both work out to
[a²+b²+c²-ac+(c-a)(√(a²+b²)+√(b²+c²)]/[c-a+√(a²+b²)+√(b²+c²)]

Edited on March 11, 2014, 7:48 am

Posted by Jer on 2014-03-10 23:45:18