IMO 1988 - Shortlisted problems Question 23

(SIN 2) Let $Q$ be the center of the inscribed circle of a triangle $ABC$. Prove that for any point $P$, $$a(PA)^2+b(PB)^2+c(PC)^2=a(QA)^2+b(QB)^2+c(QC)^2+(a+b+c)(QP)^2,$$ where $a=BC$, $b=CA$, and $c=AB$.