# Show that the bissectors of $\angle BHC$ and $\angle BFC$ meet on $BC$

euclidean-geometrygeometry

Given an acute triangle $$\triangle ABC$$ with orthocenter $$H$$. Let $$D = BH \cap AC, E = CH \cap AB$$ and $$F = (AEDH) \cap (ABC) \neq A$$. Show that the inner angle bisectors of $$\angle BFC$$ and $$\angle BHC$$ meet on the side $$BC$$.

It is easy to prove that the dotted ray passes through the midpoint of $$BC$$ and the antipode of $$A$$ in the $$(ABC)$$.

My first idea was to find $$\angle CBF$$ but that is a bit tough. So maybe drawing the perpendiculars from the meeting of the bissector of $$\angle CHB$$ with $$BC$$ to $$FB$$ and $$FC$$ may lead to something but it seems I need the forementioned angles.

We use this fact that the bisector of $$\angle BFC$$ and Perpendicular bisector of BC meet on circumcircle of triangle BFC and the bisector of $$\angle BHC$$ and Perpendicular bisector of BC meet on circumcircle of triangle BHC. These two circle have common chord BC and the meeting points are both on perpendicular bisector of BC, that means bisectors of angles $$\hat {BFC}$$ and $$\hat{BHC}$$meet at G which is on BC.