<s>[INST] How can I get the $\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=3$? 
Let the triangle $ABC$ be inscribed in a circle, let $P$ denote the centroid of the triangle and let $O$ denote the circumcenter. Suppose that $A,B,C$ have coordinates $(0,0),(a,0)$ and $(b,c)$ respectively.

a) Express the coordinates of $P$ and $O$ in terms of $a,b,c$.
b) Extend line segments $AP,BP,CP$ to meet the circle in points $D,E$,and $F$ respectively. Show that 
    $$\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=3\,.$$


I have done in part a, but for part b, since $$CP\cdot PF=BP\cdot PE=AP\cdot PD=R^2-x^2\,,$$ so $$PD=\frac{CP\cdot PE}{AP}\,,$$ $$PE=\frac{CP\cdot PE}{BP}\,,$$
and $$PF=\frac{BP\cdot PE}{CP}=\frac{CP\cdot PE}{CP}\,.$$
Hence,
$$\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=\frac{AP^2}{CP\cdot PE}+\frac{BP^2}{CP\cdot PE}+\frac{CP^2}{CP\cdot PE}=\frac{AP^2+BP^2+CP^2}{CP\cdot PE}$$
How to get $$\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=\frac{AP^2+BP^2+CP^2}{R^2-x^2}\,?$$
 [/INST] I am certain that the theorem below has been proven on this site.  Unfortunately, I could not find it.

Theorem. Let $ABC$ be a triangle with circumcenter $O$ and centroid $G$.  If $R$ is the circumradius and $P$ is an arbitrary point on the plane (or in the space), then
  $$\begin{align}PA^2+PB^2+PC^2&=GA^2+GB^2+GC^2+3\,PG^2\\&=3\,\left(R^2-OG^2+PG^2\right)\,.\end{align}$$  Furthermore, the value $PA^2+PB^2+PC^2$ is minimized if and only if $P=G$, and this minimum value is $$GA^2+GB^2+GC^2=3\,\left(R^2-OG^2\right)\,.$$ 

Without loss of generality, suppose $O$ is at the origin.  Identify each point $T$ on the plane as the vector $\overrightarrow{OT}$.  We have
$$G=\frac{A+B+C}{3}\text{ and }\|A\|=\|B\|=\|C\|=R\,.$$
That is,
$$\begin{align}PA^2+PB^2+PC^2&=\left\|\overrightarrow{PA}\right\|^2+\left\|\overrightarrow{PB}\right\|^2+\left\|\overrightarrow{PC}\right\|^2
\\&=\|A-P\|^3+\|B-P\|^2+\|C-P\|^2
\\&=\|A\|^2+\|B\|^2+\|C\|^2-2\,(A+B+C)\cdot P+3\,\|P\|^2
\\&=\|A\|^2+\|B\|^2+\|C\|^2-2\,(3\,G)\cdot P+3\,\|P\|^2
\\&=\|A\|^2+\|B\|^2+\|C\|^2-3\,\|G\|^2+3\,\|G-P\|^2
\\&=3\,R^2-3\,\left\|\overrightarrow{OG}\right\|^2+3\,\left\|\overrightarrow{PG}\right\|^2=3\,\left(R^2-OG^2+PG^2\right)\,.\end{align}$$
When $P=G$, we recover the identity
$$GA^2+GB^2+GC^2=3\,(R^2-OG^2)\,,$$
whence we may also write
$$PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3\,PG^2\,.$$
Remark. The theorem above also holds in general.  There is also an integral version (which physicists call it the Parallel Axis Theorem).  

Theorem.  For a positive integer $n$, let $A_1$, $A_2$, $\ldots$, $A_n$ be $n$ points in a Euclidean space, whose barycenter (centroid) is $G$.  Then,
  $$\begin{align}\sum_{i=1}^n\,PA_i^2&=\sum_{i=1}^n\,GA_i^2+n\,PG^2\,.\end{align}$$  Furthermore, the value $\sum\limits_{i=1}^n\,PA_i^2$ is minimized if and only if $P=G$.  

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