Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.
Školjka 
has a right angle at 
. Let 
be the point on line 
such that 
and 
lies between 
. Point 
is chosen so that 
and 
is the bisector of 
. Point 
is chosen so that 
and 
is the bisector of 
. Let 
be the midpoint of 
be the point such that 
is a parallelogram. Prove that 
and 
are concurrent.