Consider the following diagram : 
I = Incenter of a triangle ABC
P = Ex-center of a triangle ABC
AD, BR, CQ are the internal angle bisectors and BP and CP are external angle bisectors.
I had proved the following using angle bisector theorem \begin{align} \frac{AI}{ID} = \frac{AC + AB}{BC}\end{align}
It is given in my book that \begin{align} \frac{AP}{DP} = \frac{AC + AB}{BC}\end{align} since they are harmonic conjugate of each other.
I need the proof for this. I have tried several ways but I am not able to get it.