For $\triangle ABC$ with circumradius $R$, orthocenter $H$, and nine-point center $O_9$, show $O_9A^2+O_9B^2+O_9C^2+O_9H^2=3R^2$. 
Prove that, if $O_9$ is the nine-point center of triangle $ABC$ with $H$ as its orthocenter, then
  $$O_9A^2+O_9B^2+O_9C^2+O_9H^2=3R^2\,.$$

What I tried to do draw $O$ as $O_9$ is the midpoint of $OH$ and tried stuff as $AO=BO=CO=R$.
 A: Let $\Delta ABC$ be a triangle with centroid $G$, then for any point $P$ we have the relation
$$ 
\underbrace{PA^2 + PB^2 + PC^2}_{\sum PA^2} 
- 3PG^2
= GA^2 + GB^2 + GC^2  \ .
$$
We write this relation for $O$ and $O_9$, thus getting
$$ \sum O_9A^2-3O_9G^2 = \sum OA^2-3OG^2 = 3R^2-3OG^2\ . $$
Now note that the points $H,G,O,O_9$ are placed on the Euler line as follows: 

Explicitly $O_9$ is the mid point of the segment $OH$, 
(since the projection of $O_9$ on each side is the mid point of the projected segment, which is a chord in the Euler circle), $G$ is placed in the proportion $GH:GO=2:1$ (same side projection argument), so $O_9G=\frac 13 O_9O=\frac 13 O_9H$. 
This gives
$$ 3OG^2 - 3O_9G^2
=
3\left(\frac 23\right)^2O_9H^2
-3\left(\frac 13\right)^2O_9H^2
=
3\left(\frac 43-\frac 13\right)O_9H^2
=
O_9H^2\ .
$$
Putting all together, we get $\sum O_9A^2+O_9H^2=3R^2$.
$\square$
A: Write $N$ for the nine-point center of the triangle $ABC$, and $O$ for the circumcenter. 
 Let $M_a$, $M_b$, and $M_c$ be the midpoints of $BC$, $CA$, and $AB$, respectively.  Use Stewert's Theorem or Law of Cosines to establish that
$$NB^2\cdot M_aC-NM_a^2\cdot BC+NC^2\cdot BM_a-M_aC\cdot BC\cdot BM_a=0\,.$$
Since $M_aC=BM_a=\dfrac12BC$, we have
$$NM_a^2=\frac{NB^2+NC^2}{2}-\frac14BC^2\,.$$
However, $N$ is the circumcenter of the triangle $M_aM_bM_c$, which is similar to the triangle $ABC$ via a homothethy centered at the centroid $G$ of the triangle $ABC$ with the scaling factor $-\dfrac12$.  Therefore, $NM_a$ is the circumradius of the triangle $M_aM_bM_c$, which is half the circumradius of the triangle $ABC$, namely, $NM_a=\dfrac12 R$.  This gives
$$\frac{R^2}{4}=\frac{NB^2+NC^2}{2}-\frac14BC^2\,.$$
We also have two similar equations:
$$\frac{R^2}{4}=\frac{NC^2+NA^2}{2}-\frac14CA^2$$
and
$$\frac{R^2}{4}=\frac{NA^2+NB^2}{2}-\frac14AB^2\,.$$
Combining all three of them yields
$$NA^2+NB^2+NC^2-\frac{1}{4}(BC^2+CA^2+AB^2)=\frac{3R^2}{4}\,.$$
From this past question,
$$BC^2+CA^2+AB^2=9R^2-OH^2\,.$$
Therefore,
$$NA^2+NB^2+NC^2-\frac14(9R^2-OH^2)=\frac{3R^2}{4}\,,$$
which is equivalent to
$$NA^2+NB^2+NC^2-\frac{OH^2}{4}=3R^2\,.$$
By Euler's Theorem, $N$ bisects $OH$; therefore, $\dfrac{OH}{2}=NH$, making
$$NA^2+NB^2+NC^2-NH^2=3R^2\,.$$
