My proof. On the complex plane, let $(O)$ be the unit circle with $A(1)$, $B(b)$, and $C(c)$. Then the orthocenter $H$ has the coordinate $h = 1 + b + c$. From $E$ and $F$, the midpoints of $AB$ and $AC$, respectively, we have $e = \frac{b+1}{2}$ and $f = \frac{c+1}{2}$. Let $D$ be the intersection of $AH$ and $(O)$, giving $d = -bc$. 1) Let $K$ be the intersection of the line passing through $H$ perpendicular to $HF$ and $BC$. Then: \[ k = \frac{2c^2 + 3bc + b + 2c}{c - 1}. \]Let $Y$ be the intersection of $DK$ and $(O)$. Then: \[ y = -\frac{-bc + b + 2c}{2b + c + 1}. \]Let $J$ be the intersection of $BY$ and the perpendicular bisector of $BK$. Then: \[ j = -\frac{-bc^2 + 4c^2 + 6bc + b + 4c}{c^2 - 2c + 1}. \]From this, we calculate the cross ratio: \[ \frac{y-j}{f-j}:\frac{y-o}{f-o} = \frac{y-j}{f-j} \cdot \frac{f}{y} = -\frac{\left(c^2 + 6c + 1\right)\left(2b + c + 1\right)\left(bc + b + 2c\right)}{2\left(c + 1\right)\left(b^2c^2 + 6b^2c + b^2 + 8bc^2 + 8bc + c^3 + 6c^2 + c\right)}. \]It is evident that this ratio remains constant when $b, c$ are replaced by $\bar{b} = \frac{1}{b}$ and $\bar{c} = \frac{1}{c}$, which implies it is real. Thus, $J, O, F, Y$ are concyclic. 2) Similarly, calculating $k$ and $y$, we find: \[ l = \frac{2b^2 + 3bc + 2b + c}{b - 1}, \quad z = -\frac{bc + 2b + c}{b + 2c + 1}. \]Let $M$ be the intersection of $BZ$ and $OE$. Then: \[ m = -\frac{b + c}{c + 1}. \]Let $N$ be the intersection of $CY$ and $OF$. Then: \[ n = -\frac{b + c}{b + 1}. \]Let $T$ be the intersection of $MN$ and $YZ$. Then: \[ t = \frac{-b^2c^2 - b^2c + b^2 - bc^2 + bc + c^2}{\left(b + c\right)\left(bc - 1\right)}. \]Let $P$ be the intersection of $BY$ and $CZ$. Then: \[ p =-\frac{2(b^2c^2 + 3b^2c + b^2 + 3bc^2 + 3bc + c^2)}{\left(c + 1\right)\left(b + 1\right)\left(b + c\right)}. \]Instead of proving that the line through $T$ perpendicular to $OA$ bisects $AP$, consider $S$, the reflection of $A$ about $T$. We will prove that $SP \perp OA$. Indeed, if $S$ is the reflection of $A$ about $T$, then: \[ s = 2t - a = \frac{\left(b + c - 2bc\right)\left(bc + 2b + 2c + 1\right)}{\left(b + c\right)\left(bc - 1\right)}. \]It suffices to check that $\frac{s-p}{a}$ is purely imaginary. We have: \[ \frac{s-p}{a} = \frac{b^3c^2 + b^3c + 2b^3 + b^2c^3 - 2b^2c^2 - 4b^2c + b^2 + bc^3 - 4bc^2 - 2bc + b + 2c^3 + c^2 + c}{\left(c + 1\right)\left(b + 1\right)\left(b + c\right)\left(bc - 1\right)} = -a(\bar{s}-\bar{p}). \]The proof is complete.

Here is a synthetic solution for part (b). First we show that $MN \parallel BC$. Let $AD \cap BC \equiv J$. Note that $\triangle BEM \sim \triangle HJL$ and $\triangle BEO \sim \triangle AJC$, we get $$ \frac{EM}{EO} = \frac{EM}{BE} \cdot \frac{BE}{EO} = \frac{JL}{JH} \cdot \frac{JA}{JC}. $$Let $EF \cap AH \equiv G$. Note that $\triangle HLC \cup J \sim \triangle EHA \cup G$ we see that $$ \frac{EM}{EO} = \frac{GH}{GA} \cdot \frac{JA}{JH}.$$Compute the ratio $\frac{FN}{FO}$ similarly, we get $\frac{EM}{EO} = \frac{FN}{FO}$. It yields that $MN \parallel EF \parallel BC$. Now it is suffice for us to prove the following generalization. Generalization. Given $\triangle ABC$ with circumcircle $(O)$. Let $M, N$ be points on the perpendicular bisector of $AB, AC$, respectively, such that $MN \parallel BC$. Suppose that $BM, CN$ intersect $(O)$ again at $Z, Y$, respectively. Let $YZ \cap MN \equiv T$ and $BY \cap CZ \equiv P$. $X$ is midpoint of $AP$. Prove that $TX \perp OA$. Proof. We will divide the proof into some claims: Claim 1. $BZ, CY$ and $AO$ are concurrent. Proof. Let $E, F$ be the midpoints of $AB, AC$, respectively. Let $OM \cap BC \equiv U$ and $ON \cap BC \equiv V$. Since $MN \parallel EF \parallel BC$, we get $(U, E; O, M) = (V, F; O, N)$ and it implies that $B(C, A; O, Z) = C(B, A; O, Y)$. So we can conclude that $BZ$ and $CY$ intersect at a point $I \in OA$. $\square$ The idea is clear now. Let $YZ \cap BC \equiv S$. Then we have $PS \perp \overline{O, A, I}$ by Brocard theorem. Now it is enough to prove the following claim. Claim 2. Let $R$ be the intersection of the tangent at $A$ of $(ABC)$ and the line $YZ$. Then $T$ is the midpoint of $RS$. Proof. Let $B_1, C_1$ be the symmetric point of $B, C$ with respect to $M, N$, respectively. It is enough to prove that $R, B_1, C_1$ are collinear. Let $B'$ and $C'$ be the antipodes of $B, C$ in $(ABC)$, respectively. It is easy to see that two triples of points $(A, B', B_1)$ and $(A, C', C_1)$ are collinear. By Pascal theorem for $\left( \begin{matrix} A & Y & B \\ Z & A & B' \end{matrix} \right)$ we get $R, B_1$ and $B'Y \cap AB$ are collinear. By Pascal theorem for $\left( \begin{matrix} C' & Y & B \\ C & A & B' \end{matrix} \right)$, note that $BC$ and $B'C'$ intersect at infinity, the point $B'Y \cap AB$ lies on the parallel line to $BC$ passing through $C_1$. Thus, $B'Y \cap AB \in B_1C_1.$ Combine both those above facts, we get the collinearity of $R, B_1, C_1$. $\square$ By Claim 2, we see that $TX$ is the midline of the trapezoid $ARPS$, so $TX \parallel AR$. Note that $AR$ is a tangent of $(O)$, it yields that $TX \perp OA$. The proof is done. $\blacksquare$

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More general problem. Let $\triangle ABC$ be inscribed in a circle $\omega$. Points $E$ and $F$ lie on $CA$ and $AB$, respectively, such that $EF \parallel BC$. Choose a point $P$ such that $EP \perp CA$ and $FP \perp AB$. On the lines $PF$ and $PE$, take points $M$ and $N$, respectively, such that $MN \parallel BC$. The lines $BM$ and $CN$ intersect $\omega$ again at $Y$ and $Z$, respectively. Let $S$ be the intersection of $YZ$ and $MN$. Take a point $T$ such that $FT \parallel BZ$ and $ET \parallel CY$. Prove that $ST \perp PA$.

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One problem I discover from this problem. Problem(Own). Given a cyclic quadrilateral \(ABCD\) inscribed in a circle \((O)\). The diagonals \(AC\) and \(BD\) intersect at \(P\), and \(AD\) intersects \(BC\) at \(E\). Construct points \(M\) and \(N\) on \(AD\) and \(BC\), respectively, such that \(MN \parallel CD\). Let \((ONB)\) intersect \((OMA)\) at the second point \(Q\), and let \(MN\) intersect \(AB\) at \(K\). Prove that the line passing through \(K\), perpendicular to \(EO\), passes through the midpoint of \(PQ\).

Another discover: Problem 2(Own). Given an acute-angled triangle \(ABC\) inscribed in a circle \((O)\) with center \(O\) and orthocenter \(H\). The line \(AH\) intersects \((O)\) again at \(D\), distinct from \(A\). Let \(F\) be the midpoint of the segment \(AC\). The line through \(H\) perpendicular to \(HF\) intersects \(BC\) at \(K\). The line \(DK\) intersects \((O)\) again at \(Y\), distinct from \(D\). The line \(YF\) intersects \((O)\) at a second point \(T\). The perpendicular bisector of \(BK\) intersects \(YB\) at \(J\). Prove that the circumcenter of triangle \(OJF\) lies on \(TB\).

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buratinogigle wrote: More general problem. Let $\triangle ABC$ be inscribed in a circle $\omega$. Points $E$ and $F$ lie on $CA$ and $AB$, respectively, such that $EF \parallel BC$. Choose a point $P$ such that $EP \perp CA$ and $FP \perp AB$. On the lines $PF$ and $PE$, take points $M$ and $N$, respectively, such that $MN \parallel BC$. The lines $BM$ and $CN$ intersect $\omega$ again at $Y$ and $Z$, respectively. Let $S$ be the intersection of $YZ$ and $MN$. Take a point $T$ such that $FT \parallel BZ$ and $ET \parallel CY$. Prove that $ST \perp PA$.

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In fact, for $(b)$ there is a completely projective generalisation, considering the elementary narrative I replaced the conics with circles. (for the general case we are only used to replace two circular points to arbitrary points) $\textbf{Proposition.}$ Given $\triangle$$ABC$. Assume that $E\in AC$, $F\in AB$, $EF\cap BC=T$, $P\in \odot (AEF)$, $D\in AP$, $EP\cap AB=K$, $FP\cap AC=L$, $KL,EF$ intersects the tangent of $\odot (ABC)$ at $A$ at $Y,W$ resp., $DB\cap \odot (ABC)/\{B\}=U$, $DC\cap \odot (ABC)/\{C\}=V$, $BV\cap CU=X$, $UV\cap MN=Z$, $YZ\cap AX=Q$, $EQ\cap CU=I$, $FQ\cap BV=J$, $AP\cap \odot (AWT)/\{A\}=R$, then $R\in IJ$. The proof is only some applications of cross ratio and conics. Let $EF\parallel BC$ and $P$ be the antipode of $A$ wrt. $\odot (AEF)$, we get the former generalisation.

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Here is a synthetic solution for part (a). Let $H', K', X$ be reflection of $H$ wr $AB$, intersection of $H'Y$ with $AB$ and circumcenter of $KK'B$ respectively. By Pascal theorem to $ABCH'YD$ we have that $K,H,K'$ are collinear. Using butterfly theorem on circle with diameter $AC$ gives us that $H$ is midpoint of $KK'$, but also we know $HF \perp KK'$, so $X,H,F$ are collinear (perpendicular bisector of $KK'$). Let $\beta = ABC$ (angle). Now we can see that angle $K'YK$ is equal to $\pi - 2\beta$ (it's a sum of $H'YB$ and $BYD$ and we know this angles by circle $(ABC)$), but $KXK' = 2\beta$, because $X$ is circumcenter, so $KXK'Y$ is cyclic. From that we know that $BYD=BAD=\frac{\pi}{2}-\beta =XK'K=XYK=XYK$, so $B,X,Y$ are collinear. Therefore $X$ is a point which was defined in problem. Let $\gamma = ACB$ (angle). Then by angle chasing we have that $XYO = BYO = \frac{\pi}{2}-(\gamma + ACY) = \frac{\pi}{2}-(\gamma + ADY) = \frac{\pi}{2}-(\gamma + DHK) = \frac{\pi}{2}-(\gamma +\frac{\pi}{2}-BKK') = BKK' - \gamma = \frac{\pi}{2} - EFH - \gamma = \frac{\pi}{2} - XFA = XFO$ , so $XYOF$ is cyclic.