# Line segments intersecting at the nine-point center

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.