Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it China

In an acute triangle $ABC$ , $AB > AC$ , the bisector of angle $BAC$ and side $BC$ intersect at point $D$ , two points $E$ and $F$ are in sides $AB$ and $AC$ , respectively, such that $B, C, F, E$ are concyclic. Prove that the circumcenter of triangle $DEF$ coincides with the innercenter of triangle $ABC$ if and only if $BE + CF = BC$ .

Solution

Let $I$ be the innercenter of $\triangle ABC$ .

(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 $\angle ACB$ , $\triangle BIK$ and $\triangle BIE$ are reflection with respect to $BI$ , $\triangle CIK$ and $\triangle CIF$ 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 $\triangle AEF$ meet at $I$ , $IE = IF$ . Since the bisector of $\angle EBD$ and the circumcircle of $\triangle BED$ also meet at $I$ , $IE = ID$ . So, $ID = IE = IF$ , that is, $I$ is also the circumcenter of $\triangle DEF$ .

Q. E. D.

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