Proving $a|\vec{IA}|+b|\vec{IB}|+c|\vec{IC}|=0$ where I is incentre 
Let I be the incentre of triangle ABC.Using vectors prove that for any
  point P
  $a(\vec{PA})^2+b(\vec{PB})^2+c(\vec{PC})^2=a(\vec{IA})^2+b(\vec{IB})^2+c(\vec{IC})^2+(a+b+c)(\vec{IP})^2$ where
  $a,b,c$ are the sides opposite to the vertices $A,B,C$ respectively.

Regarding this sum its written in my book that 
$a|\vec{IA}|+b|\vec{IB}|+c|\vec{IC}|=0$
And that's what I'm not able to show mathematically.Any ideas?
 A: It should be : $$a\vec{IA}+b\vec{IB}+c\vec{IC}= \vec{0}$$
Let $D=BC \cap IA$ .
We'll work with position vectors with respect to some arbitrary origin .
From angle's bisector theorem in $\triangle ABC$ : 
$$\frac{BD}{CD}=\frac{c}{b}$$
Now write this in vector form :
$$b\vec{BD}=c \vec{DC}$$
$$b(\vec{D}-\vec{B})=c (\vec{C}-\vec{D})$$
$$\vec{D}=\frac{b\vec{B}+c\vec{C}}{b+c}$$
Also it's easy to see from $\frac{BD}{CD}=\frac{c}{b}$ that $$BD=\frac{ca}{b+c}$$
Now use again angle bisector's theorem in $\triangle ABD$ :
$$\frac{AI}{ID}=\frac{c}{\frac{ca}{b+c}}=\frac{b+c}{a}$$
Now transform this again in vector form :
$$a(\vec{I}-\vec{A})=(b+c)(\vec{D}-\vec{I})$$
$$\vec{I}=\frac{a\vec{A}+(b+c)\vec{D}}{a+b+c}=\frac{a\vec{A}+b\vec{B}+c\vec{C}}{a+b+c}$$
Using this we can find the first expression:
$$a\vec{IA}+b\vec{IB}+c\vec{IC}=a(\vec{A}-\vec{I})+b(\vec{B}-\vec{I})+c(\vec{C}-\vec{I})=a\vec{A}+b\vec{B}+c\vec{C}-(a+b+c)\vec{I}=a\vec{A}+b\vec{B}+c\vec{C}-(a+b+c) \cdot \frac{a\vec{A}+b\vec{B}+c\vec{C}}{a+b+c}=a\vec{A}+b\vec{B}+c\vec{C}-(a\vec{A}+b\vec{B}+c\vec{C})=\vec{0}$$ as wanted .
