Bariz - Problem

Jafet Baca`s pdf Lemma: Let ABD be a triangle and $E$ the second point of intersection of the $B$−symmedian with its circumcircle. Then $M$, the midpoint of the chord $BE$, is the center of spiral similarity which maps $AB$ to $BD$. Proof: Line $BE$ is a symmedian of $\triangle{ABD}$, so $AD$ is a symmedian of $\triangle{EAB}$ and $\triangle{EDB}$, therefore, $\angle{MAB} = \angle{DAE} = \angle{DBE} = \angle{DBM}; \angle{BDM} = \angle{ADE} = \angle{ABE} = \angle{ABM}$ whence, $\triangle{AMB}$, $\triangle{AED}$, $\triangle{BMD}$ are similiar in the same orientation. Evidently, this special center of spiral similarity inherits the aforesaid uniqueness. The result follows. Because $BE$ is also a symmedian of $\triangle{AED}$, we can show along the same lines that $\triangle{AME}$, $\triangle{ABD}$ and $\triangle{EMD}$ are pairwise directly similar; thus, $M$ takes $AE$ to $ED$. In summary, we have deduced the following results. Conclusion is trivial.