Slovenian Mathematical Olympiad, 2016

II/3. Due to different configurations we will proceed using directed angles.
Let $T$, distinct from $E$, be the intersection of the circumcircles of the triangles $BDE$ and $ACE$. The power line (radical axis) of the circumcircles of the triangles $ABC$ and $BDE$ is $BF$, the power line of the circumcircles of the triangles $ABC$ and $ACE$ is $AC$, and the power line of the circumcircles of the triangles $ACE$ and $BDE$ is $ET$. The Power-of-a-Point theorem states that these three lines intersect at a single point. The lines $AC$ and $BF$ intersect at $P$, so $P$ must also lie on the line $ET$. Due to the concyclicity of the points $B, T, D$ and $E$ we have
$$\asymp TBA = \asymp TBE = \pi - \asymp EDT = \asymp TDC,$$
and due to concyclicity of $A, T, C$ and $E$ we have
$$\asymp BAT = \pi - \asymp TAE = \asymp ECT = \asymp DCT.$$
The triangles $ABT$ and $CDT$ have three matching angles and $|AB| = |CD|$, so they are congruent. This implies that $|AT| = |CT|$ and $|BT| = |DT|$. Since $|AB| = |BC| = |CD|$, we see that the triangles $ABT$ and $CBT$ have three matching sides and are congruent and the same is true for the triangles $BCT$ and $DCT$. These two congruences imply that $\asymp TBA = \asymp CBT$ and $\asymp TCB = \asymp DCT$.
In the triangle $BCE$ the point $T$ lies on the intersection of the bisectors of the angles $\asymp CBE$ and $\asymp ECB$. So, $T$ is the incentre of the triangle $BCE$ and the bisector of the angle $\asymp BEC$ also passes through $T$. The points $E, P$ and $T$ are collinear, so $EP$ is the bisector of $\asymp AED$.