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 $BC$.
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It will be assumed that quadrilateral $ABCD$ is convex, though some results remain valid in non-convex case as well. For the proof several additional lines and points will be needed.
Introduce the notations: $p_B=EB$, $p_C=EC$, $\theta=\widehat{AEB}$.
Observe that $\triangle A'B'P$ is similar to $\triangle ABE$ with ratio $\frac{1}{2}$. It follows: $$PP'=\frac{1}{2}EH_B=\frac{1}{2}p_B\cos\theta.\tag{1} $$
Considering the triangles $C'D'R$ and $CDE$ one similarly obtains: $$RR'=\frac{1}{2}EH_C=\frac{1}{2}p_C\cos\theta.\tag{2} $$ The negative sign in expressions (1) and (2) means that the points $P'$ and $R'$ are more distant from the point $Q$ than $P$ and $R$, respectively. In other words they lie outside of the quadrilateral.
Combining (1) and (2) results in $$ \frac{PP'}{RR'}=\frac{p_B}{p_C}.\tag{3} $$
Consider now triangles $PFQ$ and $RFQ$. By law of sines: $$ \frac{PF}{\sin\widehat{PQF}}=\frac{QF}{\sin\widehat{FPQ}}\quad\text{and}\quad \frac{RF}{\sin\widehat{RQF}}=\frac{QF}{\sin\widehat{FRQ}}. $$
Observe that the RHS of both equations are equal as $\angle FPQ=\angle FRQ$ due to the fact that points $P,R,Q,F$ lie on the same circle. Thus $$ \frac{PF}{RF}=\frac{\sin\widehat{PQF}}{\sin\widehat{RQF}} =\frac{\sin\widehat{ECB}}{\sin\widehat{EBC}}=\frac{p_B}{p_C},\tag{4} $$ the last equality following from the law of sines for triangle $EBC$. Together with observation $\angle PFR=\angle PQR=\angle BEC$ this gives as a side result the similarity relation $\triangle PFR\sim\triangle BEC$.
Combining (3) and (4) one obtains: $$ \frac{PP'}{RR'}=\frac{PF}{RF}, $$ which together with already mentioned fact that $\angle FPQ=\angle FRQ$ implies: $$ \triangle FPP'\sim\triangle FRR'. $$
Particularly it means: $$ \angle PP'F=\angle RR'F\Rightarrow \angle FP'Q=\angle FR'Q\tag{5}. $$ The last equality implies that points $P',R',Q,F$ lie on the same circle. By construction it is the circle with diameter $QO$ ($O$ being the intersection of lines ${\cal B}_{AE}$ and ${\cal B}_{DE}$), and the conclusion $(OF)\perp(QF)$ immediately follows. QED.
The above proof is valid quite generally with two exceptions.
In the case of orthodiagonal quadrilaterals ($\cos\theta=0$) the points $P'$ and $R'$ coincide with $P$ and $R$, respectively. Thus $O$ lies on the circumscribed circle of $\triangle PQR$ and the claim follows.
The second exception concerns the case when $Q$ and $F$ coincide, so that the circumscribed circle of $\triangle PQR$ is tangent to $(BC)$. It can be easily demonstrated that this is possible if and only if the quadrilateral $ABCD$ is isosceles trapezoid ($BC\parallel AD$, $AB=CD$). As in this case the perpendicular bisectors to $\overline{AD}$ and $\overline{BC}$ coincide the claim is valid as well.
A comment on equation (5) is still needed. If one of the inequalities: $$ p_B\cos\theta>AC,\quad\text{or}\quad p_C\cos\theta>BD, $$ is fulfilled (the inequalities cannot be fulfilled simultaneously), the corresponding point $P'$ or $R'$ will lie below the line $(BC)$. Let it be $P'$. Then instead of (5) one obtains: $$ \angle PP'F=\angle RR'F\Rightarrow \angle FP'Q=\pi-\angle FR'Q,\tag{5'} $$ and the conclusion that points $P',R',Q,F$ are concyclic remains valid.
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Let a triangle $\triangle{TBC}$; choosing in sides $\overline{TB}$ and $\overline{TC}$ points $A$ and $D$ respectively we determine a (convex) quadrilateral $ABCD$. Take $B(0,0)$, $C(2c,0)$, $A(2a_1,2a_2)$ and $D(2d_1,2d_2)$ so that $P(a_1,a_2)$,$Q(c,0)$, $R(d_1+c,d_2)$.
Point $E$ is not arbitrary, it is the intersection of lines $\overline{AC}$ and $\overline{BD}$.
Now the verification of the property is direct but tedious as follows:
(1) Calculate the abscissa $x_O$ of the circumcenter of triangle $\triangle{AED}$.
(2) Calculate the circumcircle of triangle $\triangle{PQR}$; let $(x-\alpha)^2+(y-\beta)^2=r^2$ its equation (where $(\alpha,\beta)$ is the circumcenter of the triangle and $r$ the distance of it to one of the three vertices).
(3) Making $y=0$ in this equation we get two values $x=\alpha\pm\sqrt{r^2-\beta^2}$. The fact that $x=\alpha-\sqrt{r^2-\beta^2}$ be equal to the abscissa $x_O$ of the first step 1) is clearly equivalent to the required perpendicularity (if not would be so then the property is not true).
NOTE.-With numerical values one can verify the property easily enough taking $r=\dfrac{abc}{\sqrt{p(p-a)(p-b)(p-c)}}$ where $a,b,c$ are the sides and $p$ the semiperimeter.
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I rotated the given diagram a bit so that I can make efficient use of space to give a larger picture.
We first assume that $\angle QFX = 90^0$ (where X is point on the circle PQR whose center is at $K$). Note that QKX is then a diameter.
Extend PK to cut the red circle at T such that PKT is another diameter with $\angle PQT = 90^0$.
By midpoint theorem, the green shaded part is a //gm. From which, we get $\alpha = \alpha’ = \alpha’’$. In addition, if QT cuts EC at S, then $\angle QSC = 90^0$.
Draw EY//QT cutting the blue circle at Y. Then, AY is a diameter of the blue circle (because $\angle AEY = 90^0$). Let FX extended cut AY at some point called Z.
Claim:- Z (or Z', see later text) is $O$, the center of circle ADE. This is done if we can show that (1) $\angle AZ(or Z')D = 2 \times \angle AED$; and (2) $\triangle AZ(or Z')D$ is isosceles.
Note that (1) $\theta = \theta_1 = \theta_2 = \theta_3$; (2) $\theta_3 = \theta_4$; and (3) $\angle AEY = \theta_1 + \alpha’’ = 90^0$.
Draw a perpendicular through E cutting the blue circle at V. When DV is joined, it will cut AY at some point called Z’. Then, $\theta’’ = \theta’ = \theta_1 = \theta$.
This proves $\triangle Z'DA \sim \triangle KRP$. Therefore, $\angle Z'DA = \angle KRP$. Therefore, by angle sum of triangle, $\angle AZ'D = \angle PKR = 2 \alpha = ... = 2 \alpha’’$. Also, $\theta = \theta''$ means the triangle in question is isosceles.
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 $OF⊥BC$.