How to prove this geometry question on quadrilateral and circles? In quadrilateral $ABCD$, $AC$ and $BD$ crossed at point $E$. Circle $O$ passes through $A, D$, and $E$ and its center is $O$. $P, Q, R$ are the midpoint of $AB$, $BC$ and $CD$, respectively. Circle $O_2$ passes through $P, Q$, and $R$, and crosses $BC$ at $F$.
Prove: $OF$ is perpendicular to $BC$.


It suffices to prove that$$
BO^2 - CO^2 = BF^2 - CF^2. \tag{1}
$$
Denote the radius of circle $ADE$ by $r$. The power of point $B$ with respect to circle $ADE$ is$$
BO^2 - r^2 = BE \cdot BD,
$$
and the power of point $C$ with respect to circle $ADE$ is$$
CO^2 - r^2 = CE \cdot CA,
$$
also $BD = 2 QR$, $CA = 2 QP$, therefore$$
BO^2 - CO^2 = BE \cdot BD - CE \cdot CA = 2(BE \cdot QR - CE \cdot QP). \tag{2}
$$
Because $P, F, Q, R$ are concyclic and $BD$ and $QR$ are parallel, then$$
∠FPR = ∠CQR = ∠EBC.
$$
Analogously,$$
∠FRP = ∠BQP = ∠ECB.
$$
Therefore,$$
△BCE \sim △PRF,
$$
which implies$$
\frac{BE}{PF} = \frac{CE}{RF} = \frac{BC}{PR},
$$
or$$
BE = PF \cdot \frac{BC}{PR}, \quad CE = RF \cdot \frac{BC}{PR}. \tag{3}
$$
Now, there are two cases with respect to the relative positions of $F$ and $Q$.
Case 1: If segment $PQ$ and $FR$ intersect, then$$
BF^2 - CF^2 = (BF + CF)(BF - CF) = BC \cdot (-2QF) = -2BC \cdot QF. \tag{4}
$$
Therefore,\begin{align*}
(1) &\stackrel{(2)(4)}{\Longleftrightarrow} BE \cdot QR - CE \cdot QP = -BC \cdot QF\\
&\stackrel{(3)}{\Longleftrightarrow} PF \cdot \frac{BC}{PR} \cdot QR - RF \cdot \frac{BC}{PR} \cdot QP = -BC \cdot QF\\
&\Longleftrightarrow PF \cdot QR - RF \cdot QP = -PR \cdot QF\\
&\Longleftrightarrow PF \cdot QR + PR \cdot QF = RF \cdot QP,
\end{align*}
and the last equality holds due to Ptolemy's theorem.
Hence (1) holds, which implies $OF⊥BC$.
Case 2: If segment $PQ$ and $FR$ do not intersect, then$$
BF^2 - CF^2 = (BF + CF)(BF - CF) = BC \cdot 2QF = 2BC \cdot QF. \tag{4'}
$$
Therefore,\begin{align*}
(1) &\stackrel{(2)(4)}{\Longleftrightarrow} BE \cdot QR - CE \cdot QP = BC \cdot QF\\
&\stackrel{(3)}{\Longleftrightarrow} PF \cdot \frac{BC}{PR} \cdot QR - RF \cdot \frac{BC}{PR} \cdot QP = BC \cdot QF\\
&\Longleftrightarrow PF \cdot QR - RF \cdot QP = PR \cdot QF\\
&\Longleftrightarrow RF \cdot QP + PR \cdot QF = PF \cdot QR,
\end{align*}
and the last equality holds due to Ptolemy's theorem.
Hence (1) holds, which implies $OF⊥BC$.
