Proof 1: Let $O$ and $H$ be the circumcenter and orthocenter of $\triangle ABC$, respectively, and let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. Choose three points $A_3$, $B_3$, and $C_3$ on circle $\Gamma$ such that $AA_3$, $BB_3$, and $CC_3$ are diameters of circle $\Gamma$ (as shown in Figure 1).

Since $A_3B \perp AB$, we have $A_3B \parallel CH$.
Similarly, $A_3C \parallel BH$.
Thus, we obtain quadrilateral $A_3CHB$ and $D$ is the midpoint of $HA_3$.
Similarly, $E$ and $F$ are the midpoints of $HB_3$ and $HC_3$, respectively.
Let the orthogonal projections of point $O$ on lines $PA$, $PB$, and $PC$ be $A_4$, $B_4$, and $C_4$, respectively. Then points $A_4$, $B_4$, and $C_4$ all lie on the circle with diameter $OP$, denoted as $\Gamma_1$.

Since $D$ is both the midpoint of $HA_3$ and $A_1A_2$, we have $HA_2 = A_1A_3$.

Also, in right triangles $\triangle AA_1A_3 \sim \triangle AA_4O$, and $\frac{AO}{AA_3} = \frac{1}{2}$, we have $A_1A_3 = 2A_4O$. Therefore,
$HA_2 = A_1A_3 = 2OA_4$.
Similarly, $HB_2 = 2OB_4$, $HC_2 = 2OC_4$.
Thus, there exists a homothety that maps points $(H, A_2, B_2, C_2)$ to $(O, A_4, B_4, C_4)$.

Since $O$, $A_4$, $B_4$, and $C_4$ lie on circle $\Gamma_1$, points $A_2$, $B_2$, $C_2$, and $H$ are concyclic.

It is worth noting that a simpler proof can be obtained using vectors.

Proof 2: Let $D$, $E$, and $F$ be the midpoints of $BC$, $CA$, and $AB$, respectively. Choose point $X$ such that $\boldsymbol{OX} = \boldsymbol{PH}$.

We will prove that $|XB_2| = |XH|$ (i.e., point $X$ lies on the perpendicular bisector of $HB_2$, and similarly, point $X$ lies on the perpendicular bisectors of $HC_2$ and $HA_2$, thus, point $X$ is the center).
Notice that
$$\begin{array}{l} OX = PH = OH - OP \\ = OA + OB + OC - OP. \end{array}$$
(Using $OH = 3OG$ (Euler line) $= OA + OB + OC$)
$$\begin{array}{l} \text{Also, } OA_2 = OA_1 + A_1A_2 = OA_1 + 2A_1D \\ = OA_1 + 2\left(OD - OA_1\right) \\ = OA_1 + 2\left(\frac{OB + OC}{2} - OA_1\right) \\ = OB + OC - OA_1. \end{array}$$

Similarly, $OB_2 = OA + OC - OB_1$.
Subtracting (2) from (1), we get
$$\begin{array}{l} B_2X = OX - OB_2 = OB + OB_1 - OP \\ = OB + PB_1 = OB + BP' = OP', \end{array}$$
where point $P'$ lies on $BB_1$ such that $\boldsymbol{BP'} = \boldsymbol{PB_1}$.
Thus, $|B_2X| = |OP'| = |OP|$.
But quadrilateral $OXHP$ is a parallelogram, so
$|OP| = |XH|$.
Therefore, $|XB_2| = |XH|$.
Additionally, it is easy to find variations of this problem, such as changing "P is an interior point of $\triangle ABC$" to "P is any point in the plane of $\triangle ABC$", and discussing how the conclusion changes; or changing "the symmetric points about the midpoints of sides $BC$, $CA$, and $AB$" to "the symmetric points about sides $BC$, $CA$, and $AB$", or to "the symmetric points about the circumcenter $O$ of $\triangle ABC$", and conducting similar research.