Can you provide a proof for the following claim:
Claim.Given any $\triangle ABC$. The tangent lines to the circumscribed circle of $\triangle ABC$ are constructed at vertices $A$,$B$,$C$ . Let point $D$ be the intersection point of tangent lines that passes through vertices $A$ and $C$ , point $E$ the intersection point of tangent lines that passes through vertices $A$ and $B$ , point $F$ the intersection point of tangent lines that passes through vertices $B$ and $C$ , and let $H_1$,$H_2$,$H_3$ be the orthocenters of $\triangle ACD$, $\triangle AEB$ and $\triangle BFC$ respectively. Then line segments $AH_3$,$BH_1$ and $CH_2$ concur at the nine-point center of $\triangle ABC$.
GeoGebra applet that demonstrates this claim can be found here.
I don't know how to start the proof. All I know is that nine-point center of the triangle lies in the middle of the line segment whose endpoints are orthocenter and circumcenter , but I don't know how to use that fact. Any hints are welcomed.
In fact we can say more : $N_9$ is the midpoint of segments $AH_i$.
First we show that $BOCH_2$ is a rhombus. $\triangle BEC$ is isosceles so $H_2$ lies on its symmetry axis. So $BH_2C$ is isosceles as is $BOC$. Hence $OH_2$ is perpendicular to $BC$ at its midpoint $M$. Also $\angle OBC = 90 -A = 90 - \angle EBC = \angle H_2CB$.
Next we show $AHH_2O$ is a parallelogram. $AH \perp BC \Rightarrow AH || OH_2$. It is known that $2OM = AH$. So $AH = OH_2$.
We conclude $AH_2$ is bisected at intersection of diagonals of parallelogram $AHH_2O$. But midpoint of $OH$ is $N_9$. We're done.