CatalinBordea wrote: Let $ ABC $ be an acute triangle having $ AB<AC, $ let $ M $ be the midpoint of the segment $ BC, \color {red}{D}$ be a point on the segment $ AM, E $ be a point on the segment $ BD $ and $ F $ on the line $ AB $ such that $ EF $ is parallel to $ BC, $ and such that $ AE $ and $ DF $ pass through the orthocenter of $ ABC. $ Prove that the interior bisectors of $ \angle BAC $ and $ \angle BDC, $ together with $ BC $ are concurrent. Vlad Robu Did you mean this ?

Wiz-of-Oz wrote: CatalinBordea wrote: Let $ ABC $ be an acute triangle having $ AB<AC, $ let $ M $ be the midpoint of the segment $ BC, \color {red}{D}$ be a point on the segment $ AM, E $ be a point on the segment $ BD $ and $ F $ on the line $ AB $ such that $ EF $ is parallel to $ BC, $ and such that $ AE $ and $ DF $ pass through the orthocenter of $ ABC. $ Prove that the interior bisectors of $ \angle BAC $ and $ \angle BDC, $ together with $ BC $ are concurrent. Vlad Robu Did you mean this ? Yes.

Here's my solution: Note that it suffices to show that $D$ lies on the $A$-Apollonius circle of $\triangle ABC$, or equivalently that $D$ is the $A$-Humpty point of $\triangle ABC$. So let us assume that $D$ is the $A$-Humpty point and $H$ be the orthocenter of $\triangle ABC$. Denote $F=AB \cap HD$ and let $E$ be the point on $AH$ such that $EF \parallel BC$. We will show that $DE$ passes through $B$. Note that $\angle FEA=\angle FDA=90^{\circ}$, which means that $ADEF$ is cyclic. Thus, $$\angle EDF=\angle EAF=\angle BAH=\angle BCH=\angle BDH \Rightarrow B,E,D \text{ are collinear. } \blacksquare$$