China Mathematical Olympiad

Let $I$ be the innercenter of $△ABC$.

(Sufficiency) Suppose $BC=BE+CF$. Let $K$ be the point on $BC$ such that $BK=BE$, thus $CK=CF$. Since $BI$ bisects $\angle ABC$, $CI$ bisects $\angle ACB$, $△BIK$ and $△BIE$ are reflection with respect to $BI$, $△CIK$ and $△CIF$ are reflection with respect to $CI$, we have $\angle BEI=\angle BKI=\pi -\angle CKI=\pi -$

$\angle CFI=\angle AFI$. Therefore, $A,E,I,F$ are concyclic. Since $B,E,F,C$ are concyclic, we have $\angle AIE=\angle AFE=\angle ABC$, and hence $B,E,I,D$ are concyclic.



Since the bisector of $\angle EAF$ and the circumcircle of $△AEF$ meet at $I$, $IE=IF$. Since the bisector of $\angle EBD$ and the circumcircle of $△BED$ also meet at $I$, $IE=ID$. So, $ID=IE=IF$, that is, $I$ is also the circumcenter of $△DEF$.

Q. E. D.