How can I get the $\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=3$?

Question

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 $$\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=\frac{AP^2+BP^2+CP^2}{R^2-x^2}\,?$$

Answer 1

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 $P=G$.