Geometry, points on a circumcircle, perpendicularity Consider an acute triangle ABC with midpoints D,E,F of sides BC, CA, AB respectively. Let CF intersect the perpendicular bisectors of AC and BC at M and N respectively. Now, consider the point of intersection X of the lines AM and BN.
Moreover, let K be the intersection of CX and the circumcircle of ABC and let H be the intersection of BX and the circumcircle of ABC. Let g be the line through K that is parallel to AC. Let J be the intersection of g with the circumcircle of ABC.
Prove that OC is perpendicular to HJ.
This is a part of the problem discussed here: Equal angles geometry questions.
Somebody provided an incomplete proof for the original problem; however the part described above is still missing.
 A: 
We want to prove that $CJ=CH$, but since $CJ=AK$ it is enough to show that $CK\parallel AH$.
Notice that the perpendicular to $CA$ through $A$ and the perpendicular to $CB$ through $B$ meet on the circumcircle at the antipode $C'$ of $C$: by considering the quadrilaterals $CAC'B$ and $CAC'K$ it follows that $X$ is the midpoint of $CK$ and $OX$ is the perpendicular bisector of $CK$. Additionally  $AK=2 EX$. Since $OXCE$ is a cyclic quadrilateral, $\widehat{EXO}=\widehat{ECO}=\widehat{ACO}=\frac{\pi}{2}-\widehat{B}$ and $\widehat{CXE}=\widehat{B}$. The problem reduces to showing that $\widehat{HCK}=\widehat{HBK}=\widehat{B}$. I achieved this in a convoluted way, i.e. by introducing the antipode $B'$ of $B$, the intersection $L=AX\cap \Gamma$ and by checking that $B'L\parallel OX$. It can probably be done in more efficient ways, but I guess I have at least outlied some ideas.
A: $\angle 1 = \angle 2 = \angle =3$ implies  (arc FC) = (arc CF).

That is, since OC divides arc HJ equally, OC will also divide chord HJ equally. Result follows.
