Let $M$ be the midpoint of $BC$, $G$ the symmetric of $D$ wrt to $M$, $\{X\}=FM\cap (ABC)$ and $\{Y\}=XD\cap EF$. As $(B,C,E,D)=-1$, $MG\cdot ME=MD\cdot ME=MB^2=MX\cdot MF$, hence $EXGF$ is cyclic. From $\widehat{XDM}=\widehat{XGM}=\widehat{XFE}$, we get that $DMFY$ is cyclic whence $XD\perp EF$. As $XDFD^\prime$ is a parallelogram, we infer that $D^\prime F \perp EF$. As a sidenote, $\widehat{XD^\prime F}=\widehat{XDF}=\widehat{XGF}$, so $D^\prime \in (EXGF)$.

with center $T.$ According to the homothety $\mathbf{H}\left(D, \tfrac{1}{2}\right)$, it is sufficient to show that $\angle OMT = 90^{\circ}.$ As was shown in the hidden text above, $\Gamma$ and $\omega$ are orthogonal. Therefore, the circle of diameter $\overline{OT}$ is just the midcircle of $\Gamma$ and $\omega.$ But as $M$ is the midpoint of $\overline{DF}$, it is clear that $M$ lies on said midcircle, so $\angle OMT = 90^{\circ}$ as desired.

My solution: Let $E'$ the reflection of $E$ in $O$ and $F'$ the reflection of $F$ in $O$ $\Longrightarrow$ $F',A,E$ are collinear and $F'\in \odot(ABC)$ Let $X$ be a midpoint of $BC$ and let $X'$ the reflection of $X$ in $O$ $\Longrightarrow$ $X'=E'D'\cap OX$ $\Longrightarrow$ $\triangle F'X'E'\cong \triangle FXE$ $\Longrightarrow$ $\angle F'E'X'=\angle FEX$ and $\angle DF'X=\angle FEX$ since $D$ is orthocenter of $\triangle FF'E$ $\Longrightarrow$ $\angle DF'X=\angle F'E'X'$ $\Longrightarrow$ $\angle DF'E'=90^{\circ}$ since $\angle X'F'E'=90^{\circ}-\angle DF'X=90^{\circ}-\angle F'E'X'$ $\Longrightarrow$ $\angle D'FE=90^{\circ}$ since $\angle D'FE$ is the reflection of $\angle DF'E'$ in $O$...

My solution: Let $EA$ intersect circle around $ABC$ again at $H$. Because the angle bisector and side bisect intersect circumcircle in the same point we have that $F$ and $H$ are reflections in $O$. Now we have that $AD\perp AE$ $\implies$ $D'H\perp EA$. Also $\angle ADE=\angle AHF$ $\implies$ $\frac{AF}{AH}=\frac{EA}{AD}$ $\implies$ $\frac{EA}{AF}=\frac {AD}{AH}$ $\implies$ $\triangle FAE$$\sim$$\triangle ADH$ $\implies$ $\angle ADH=\angle AFE$. Because $FDHD'$ is parallelogram we have $\angle D'FD=\angle D'HD=90-\angle DHA=90-\angle DFE=90-\angle AFE$ $\implies$ $\angle EFA+\angle AFD'=\angle EFD'=90$. Done.

Let $\omega$ be the circumcircle of $\triangle ABC$, and reflect $F$ in $O$ to $F'$, the midpoint of the arc $BAC$. Then $DF'\parallel D'F$ so we are to show that $DF'\perp EF$. Let $DF'$ intersect $\omega$ at $P\neq F'$. Since $FF'$ is a diameter of $\omega$ we have $\angle FPF'=90^{\circ}$ so we have to show that $\angle EPF' = 90^{\circ}$. Clearly $FF'\perp BC$ so we'll show that $F'$, $E$, $P$ and $M=BC\cap FF'$ are concyclic. But the internal and external angle bisectors are perpendicular, so $FMAE$ are cyclic, so by the power of $D$ we have $ED\cdot MD = AD\cdot FD = PD\cdot F'D$ so we are done.

My solution: Let $AE$ intersect the circumcircle of $\triangle ABC$ again at $G$, then $G$ and $F$ are obviously symmetric about $O$. Let $GD$, $EF$ intersect the circumcircle of $\triangle ABC$ again at $X_1, X_2$, respectively. Now $((B,C),(D,E))$ are harmonic, hence (projecting from $G$) $((B,C),(X_1,A))$ are harmonic, as well as (projecting from $F$) $((B,C),(A,X_2))$. Thus we must have $X_1=X_2$, and since now $GD \parallel FD'$ and $GD \perp EF$ , we conclude that $D'F \perp EF$.

Let $X$ be the point on $AD$ such that $(A,F;D,X)$ is harmonic. Let $D''$ be the point on $OD$ such that $\angle D''FE=90^{\circ}$. Since $(B,C;D,E)$ is harmonic, $XE$ is the polar of $D$, so $XE\perp OD$. Now since $E(A,F;D,X)=(EA,EF;ED,EX)$ is harmonic, by taking perpendiculars we obtain that $(FD,FD";FO, \ell)=$ is harmonic, where $\ell$ is a line parallel to $OD$. But then $\ell$ is parallel to $DD"$, so $O$ bisects $DD"$, and $D"=D'$.

$(B,C;D,E)=-1$ thus $ED.EM=EB.EC=EK.EF$ with $ K$ and $M$ are resp the intersection of the circumcircle and $EF$,the midpoint of $BC$ thus $KFDM$ is cyclic ; let $F'$ the reflection of $F$ in $O$ ,which namely the midpoint of the other arc $\overarc{BC} $ it 's clear that $AF'MD$ are cyclic then $ \widehat{D'FE}= \widehat{D'FF'}+ \widehat{F'FE}= \widehat{DF'F}+ \widehat{F'DM}=\frac{\pi}{2}$ WCP

Denote $K$ the midpoint of the arc $BAC$ is clear that $K \equiv AE \cap FO \Rightarrow DKD'F$ is a paralelogram, then $KD \parallel D'F$ Since $D$ is orthocenter of $\triangle AKF \Rightarrow EF \perp FD'$

let $F'$ and $A'$ be resp the the antipodes of $F$ and $A$ in $(ABC)$, and let $M$ be the midpoint of side $BC$ lemma: $AD.AF$=$AF'.AE$ proof: -clearly $F'$ lies on $(ABC)$ note that $EAMF$ and $AF'MD$ are cyclic,thus straightforward computations yellds: $AD.AF=AF^2-FM.FF'$ $(1)$ $AF'.AE=F'M.FF'-AF'^2$ $(2)$ therefore: $(1)=(2)$ $\Longleftrightarrow$ $AF'^2+AF^2=FF'^2$ clearly true. now, we have $\widehat{EAF}=\widehat{F'A'F}=90$, and $F'A'/FA'=AF/AF'=AD/AF'$, hence $F'A'F\sim EAF$ it follows then by angle chase $\widehat{EFD'}=90$ as desired.

Let $X$ be midpoint of $DF,$ $Y$ midpoint of $DE,$ and $M$ midpoint of $BC.$ Since $\odot(\Delta ABC)$ is Apollonius circle of $E,F$ it follows that $YA \perp AO$ and since $YM \perp OM \implies$ $AOMY$ is cyclic. Since $YAMX$ is also cyclic it follows that $AYOX$ is cyclic and thus the claim.

can any one of you send me a diagram

Let $H$ be the other end of the Diameter through $FO$ of circumcircle of $\triangle ABC$. Let $G$ be the point of intersection of line $FE$ and the circumcircle. Clearly $FH \perp$ to tangent at $F$ as $F$ happens to be the mid point of the arc $BC$. $FH$ is also $\perp BC$ as it passes through mid-point of $BC$. Thus $BC$ is parallel to tangent through $F$. So we have, by tangent-chord theorem $\angle FAG$ = tangent chord angle = $\angle DEG$ (from parallel lines), $\Rightarrow ADGE$ is cyclic. But $\angle DAE = 90^{\circ}$, so $\angle DGE = 90^{\circ}$ So $DG \perp FG \Rightarrow GD$ meets diameter $FH$ at $H$. Since $D'FDH$ forms a parallelogram, $D'F \parallel DG$ which is $\perp FE \Rightarrow D'F \perp FE$ $\Rightarrow \angle D'FE =90^{\circ}$

It was short

Maybe a bit overcomplicated solution. Let $X=FE \cap \odot (ABC) $. Observe that: $$ -1 = (E, D; B, C) \overset{F}{=} (X, A; B, C) $$Thus $X$ is intersection of $A$ symmedian in $\triangle ABC$ with $\odot(ABC)$. Let define point $M = OF \cap BC$. It is easy to see that $M$ is midpoint of $BC$ and that $OF \perp BC$. Observe that quadrilateral $EADF$ is cyclic. Thus we have: $$ \angle MAF = \angle XAF = \angle DEF $$This implies that $ADXE$ is cyclic and thus $\angle DXE = 90^{\circ}$. Consequently we have that $XD$ passes through $M_A$, where $M_A$ is midpoint of arc $BC$ containing $A$. Observe that: $$ MO = OF \quad \text{and} \quad DO = OD' $$This implies that $M_ADFD'$ is parallelogram. This ends the proof since: $$ \angle DM_AF = \angle M_AFD'= \angle FAX $$which implies that $FD'$ is tangent to $\odot(ADFE)$. Since the centre of this circle is midpoint of segment $EF$ we have that $\angle D'FE = 90^{\circ}$ as desired.

Let $N$ be the reflection of $F$ over $O$. Note that $N$ is the midpoint of arc $BAC$, i.e. $N$ lies on the external angle bisector of $\angle BAC$. Hence, we would like to show that $ND\perp EF$ as $NDO'F$ is a parallelogram. Let $K=EF\cap (ABC)$. Note that $\measuredangle FKN=\measuredangle FAN=90^\circ$. Also as $ANMD$ is cyclic, by PoP, $DMFK$ is cyclic. Thus, $\measuredangle FKD=\measuredangle FMD=90^\circ$. Hence, we conclude that $K,D,N$ are collinear and as $K$ lies on $EF$ and $\measuredangle FKD=90^\circ$, we are done. [asy][asy] size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,I,D,F,N,E,o,M,K; O=(0,0);A=dir(130);B=dir(200);C=dir(340);I=incenter(A,B,C);path w=circumcircle(A,B,C);D=intersectionpoint(B--C,2I-A--A);F=intersectionpoints(w,3I-2A--A)[1];N=-F;E=extension(A,N,B,C);o=-D;M=midpoint(B--C);K=foot(D,E,F); draw(A--B--C--cycle,deep);draw(w,med);draw(N--E,deep);draw(B--E,deep);draw(A--F,deep);draw(N--K,deep);draw(E--F,deep);draw(o--F,deep);draw(N--o,deep);draw(D--o,deep);draw(N--F,deep); dot("$O$",O,S); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$F$",F,dir(F)); dot("$N$",N,dir(N)); dot("$E$",E,dir(E)); dot("$O'$",o,dir(o)); dot("$M$",M,dir(M)); dot("$K$",K,dir(K)); [/asy][/asy]