We have: $\angle{BEF} = \angle{DEB} + \angle{FED} = 90^o - \dfrac{\angle{ABC}}{2} + 90^o - \angle{IFD} = 90^o - \dfrac{\angle{ABC}}{2} + 90^o - \angle{IFC} + \angle{CFD}$ $= 90^o - \dfrac{\angle{ABC}}{2} + 90^o - \dfrac{\angle{BAC}}{2} - \angle{ACB}$ $+ 90^o - \dfrac{\angle{ACB}}{2} = 180^o - \angle{ACB}$ Then: $B$, $C$, $F$, $E$ lie on a circle

INAMO SL 2014 G1 wrote: The inscribed circle of the $ABC$ triangle has center $I$ and touches to $BC$ at $X$. Suppose the $AI$ and $BC$ lines intersect at $L$, and $D$ is the reflection of $L$ wrt $X$. Points $E$ and $F$ respectively are the result of a reflection of $D$ wrt to lines $CI$ and $BI$ respectively. Show that quadrilateral $BCEF$ is cyclic . Solution: Notice: $\angle IEC=\angle IDL=\angle ALB \implies ECLI$ cyclic and similarly, $BFIL$ is cyclic. Hence by POP $BFEC$ is cyclic