A $\Delta ABC$ has orthocentre $H,$ nine-point centre $N$ and circumradius $R.$ For any point $P$ on its nine-point circle, prove that $$PA^2+PB^2+PC^2+PH^2=4R^2.$$
I can prove that $NA^2+NB^2+NC^2+NH^2=3R^2.$ But don’t know how to proceed further. How does the nine-point circle come into play? Any kind of help/suggestions will be highly appreciated.
On
Place the points $A,B$ and $C$ across the unit circle on the Argand plane with the coordinates $a,b$ and $c.$ Then $$|a|=|b|=|c|=R=1$$ $$H:h=a+b+c$$ $$N:n=\frac{a+b+c}{2}$$ Let $P = p.$ Since $P$ lies on the nine-point circle of $\Delta ABC,$ we have \begin{align*} & |p-n|=\frac{R}{2} \\ \implies & \left|p-\frac{a+b+c}{2}\right|=\frac{R}{2} \\ \implies & |2p-a-b-c|=R \\ \implies & (2p-a-b-c)\cdot (2p-a-b-c) = R^2 \\ \implies & 4p\cdot p + a \cdot a + b \cdot b + c \cdot c - 4p \cdot (a+b+c) +2(b \cdot c +c \cdot a + a \cdot b) = R^2 \\ \implies & 4p\cdot p + 3R^2 - 4p \cdot (a+b+c) +2(b \cdot c +c \cdot a + a \cdot b) = R^2 \;\;\; [\text{Using}\; |a|=|b|=|c|=R] \\ \implies & 4p\cdot p - 4p \cdot (a+b+c) +2(b \cdot c +c \cdot a + a \cdot b) = -2R^2 \;\;\; (1) \end{align*} Therefore \begin{align*} & PA^2+PB^2+PC^2+PH^2 \\ &= (p-a)\cdot (p-a) + (p-b)\cdot (p-b) + (p-c)\cdot (p-c) + (p-h)\cdot (p-h) \\ &= 4p\cdot p + a \cdot a + b \cdot b + c \cdot c + h \cdot h - 2p \cdot(a+b+c+h) \\ &= 4p \cdot p -4p \cdot(a+b+c) +2(a \cdot a + b \cdot b + c \cdot c) + 2(b \cdot c+ c \cdot a + a \cdot b) \;\;\; [\because h = a+b+c] \\ &= 6R^2 + \{4p \cdot p -4p \cdot(a+b+c) + 2(b \cdot c+ c \cdot a + a \cdot b)\} \;\;\; [\text{Using}\; |a|=|b|=|c|=R] \\ &= 6R^2-2R^2 \;\;\; [\text{Using}\; (1)] \\ &= \boxed{4R^2} \end{align*} $QED.$
Hint:
Let $G $ be the centroid of the triangle. Then $$ \vec{NA}+\vec{NB}+\vec{NC}=3\vec{NG}+\underbrace{\vec{GA}+\vec{GB}+\vec{GC}}_{=0}=-\vec{NH}\\ \implies \vec{NA}+\vec{NB}+\vec{NC}+\vec{NH}=0. $$ Using this and already proven result $$ NA^2+NB^2+NC^2+NH^2=3R^2 $$ the claim follows from the law of cosines.