it's enough to prove that $AFMD$ is cyclic (assume that $M$ is the midpoint of $OH$).for this it suffices to prove that $AF=MD$ and it's obvious because $AF=\frac{R}{2}$ ($R$ is the circumradii) and because $M$ is the center of ninepoints circle and $D$ is on ninepoints circle then $MD=\frac{R}{2}$ (because the radii of ninepoints circle is $\frac{R}{2}$)

I have another one: Extend $AD$ to meet $(O)$ at $S$, call $N$ as midpoint of $OH$. We have: $\widehat{OSH}=\widehat{NDH}(1)$ And $\widehat{OSA}=\widehat{OAS}(2)$ From $(1)$ and $(2)$, we have $\widehat{NDA}=\widehat{MAD}(3)$ And $MN$ is parallel to $AD(4)$ From $(3)$ and $(4)$, we have $MNDA$ is a isoceles trapezoid, so it is a cyclic quadrilateral, hence proved.

Firstly see that since $\angle AFE=\angle ADE=90^{\circ}$ we have $AFDE$ is concyclic. Again see that it will suffice to prove $AFND$ is cyclic where $N$ is the nine-point centre,i.e mid point of $OH$.Now see that $AF=ND$ since $AF$ is half of $AO$ and also $ND$ is a radius of the nine-point circle.Now it remains to be seen that $\triangle ONF$ and $\triangle OHA$ are homothetic w.r.t $O$ with ratio $2$.This gives $NF\parallel AH \implies NF\parallel AD$ hence $AFND$ is an isosceles trapezium and it is concyclic. Hence $A,F,N,D,E$ all lie on the same circle. BTW I couldn't understand the posts abovecompletely.I assume I am not posting same solutions again.

Take $P$ symmetrical of $O$ across $BC$ and $Q=OP\cap BC$. $AHPO$ is a parallelogram, $MQ$ midline in $\triangle OHP$, so $MQ=\frac{HP}2$, but $MD=MQ$, therefore $MD=AF$; with $FM\parallel AD, ADMF$ is an isosceles trapezoid, done. Best regards, sunken rock

A similar solution:By easy homotethy wrt $G$ (centroid) we see that $R(ABC)=2R$(Euler circle).Now it is trivial that $AEDF$ is cyclic.$M$ is the midpoint of $HO$ and $F$ is a midpoint of $AO$ so $MF||AD$.We have that $AF=\frac{R(ABC)}{2}=MD$ so $ADMF$ is iscosceles trapezoid so it is cyclic and the task is proven.

Dear Mathlinkers, the problem http://www.artofproblemsolving.com/community/c6h1070977_perpendicular_bisector gives a nice proof... 1. N the midpoint of OH 2. U the second point of intersection of the parallel to OH with (ADE) 3. the pencil (A; H, O, N, U) is harmonic 4. consider the pencil wit summit E perpendicular to the last pencil 5. with the problem mentioned above, we are done... Sincerely Jean-Louis

yunxiu wrote: $O$ is the circumcenter of acute $\Delta ABC$, $H$ is the Orthocenter. $AD \bot BC$, $EF$ is the perpendicular bisector of $AO$,$D,E$ on the $BC$. Prove that the circumcircle of $\Delta ADE$ through the midpoint of $OH$. I solved it just using synthetic geometry... The midpoint of $OH$ is $N$, the nine point center. Now join $MN$, also note that $MN||AH$, and $MN=0.5AH=OX$, since it is well known that $AH=2OM$,where $M$ is the modpoint of the required line segment. Also, we know that the feet of the altitudes, midpoints of the three sides, and the midpoint of the line joining the vertex to the orthocenter are concyclic. So, as the Nine point center is already $N$, hence, $NX=ND=0.5AH$(a radius of the nine point circle)$=OX=MN$. Now it is know that a quadrilateral is cyclic if two adjacent pairs are equal and the common vertex among these is the bisector of the opposite angle. Hence, it suffices to prove that either $EN$ bisects $\angle MED$, or $AN$ bisects $\angle DAM$. Certainly, proving the later part will be easier, because we do not know enough about the $\angle MED$. But it is now trivial, as $\angle BAO=\angle CAD$. This implies $AMND$ is concyclic. But as the other other points are shared by vertexes $A,M,N,D$, it implies $AMNDE$ concyclic. My second solution in AoPS.... Regards...... MD. Shamim Akhtar...$\blacksquare$

Solution (Credit - Math Pi Rate) Let $O'$ be the reflection of $O$ on $BC$ and let $N$ be the nine point center. Note that $E$ is the circumcenter of $OO'A$. Also, $N$ is the mid point of $AO'$. So, $EN\perp AO'$ and so, $A,E,N,D$ are cyclic. $\blacksquare$.

Why is $N$ the midpoint of $AO'$?

ShamimAkhtar212 wrote: Why is $N$ the midpoint of $AO'$? that is because, $AHO'O$ forms a parallelogram

AlastorMoody wrote: ShamimAkhtar212 wrote: Why is $N$ the midpoint of $AO'$? that is because, $AHO'O$ forms a parallelogram Thanks!

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 17... Sincerely Jean-Louis