Let ABCD be a parallelogram. Let the incircle of ΔABC
touch diagonal AC at point P. Let r
_{1} and r
_{2} be the inradii
of triangles APD and PCD respectively.
r_{1} AP
Prove that  = 
r_{2} 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 = (pa)/(cp)
The inradius of a circle is 2*Area/Perimeter
r1 = (pa)*b/(BC+(pa)+PD)
r2 = (cp)*b/(AB +(cp)+PD)
So for AP/PC = r1/r2 it suffices to show
BC + (pa) = AB + (cp)Now for the messy algebra
AB=√(a²+b²)
BC=√(b²+c²)
p = [c√(a²+b²)+a√(b²+c²)]/[ca+√(a²+b²)+√(b²+c²)]
The bold expressions above that must be shown to be equal both work out to
[a²+b²+c²ac+(ca)(√(a²+b²)+√(b²+c²)]/[ca+√(a²+b²)+√(b²+c²)]
Edited on March 11, 2014, 7:48 am

Posted by Jer
on 20140310 23:45:18 