Let $O$, $J$ be circumcenters of $\triangle ABC$, $\triangle AEF$ We have: $\angle{BAC} = \angle{BEA'} = \angle{CFA'} = 45^o$ Then: $EA'$ $\parallel$ $AC$, $FA'$ $\parallel$ $AB$ or $EAFA'$ is parallelogram But: $O$ is midpoint of $AA'$ so $O$ is also midpoint of $EF$ Hence: $JO$ $\perp$ $EF$ Combine with: $JO$ $\perp$ $AK$, we have: $EF$ $\parallel$ $AK$ But: $AK$ $\perp$ $KA'$ then: $EF$ $\perp$ $KA'$ or $EO$ $\perp$ $KA'$ So: $EF$ passes through midpoint of $KA'$

This problem was proposed by me Generalization: Given a triangle $ABC$ with $\angle A < 60^\circ$. Point $A'$ lies inside of $\angle A$ such that $\angle A'BA=\angle A'CA=180^\circ - 2\angle A$. Points $E$ and $F$ on segments $AB$ and $AC$ respectively are such that $A'B = BE$, $A'C = CF$. Let $K$ be the second intersection of circumcircles of triangles $AEF$ and $ABC$. Prove that $EF$ bisects $A'K$.

Dadgarnia wrote: This problem was proposed by me Generalization: Given a triangle $ABC$ with $\angle A < 60^\circ$. Point $A'$ lies inside of $\angle A$ such that $\angle A'BA=\angle A'CA=180^\circ - 2\angle A$. Points $E$ and $F$ on segments $AB$ and $AC$ respectively are such that $A'B = BE$, $A'C = CF$. Let $K$ be the second intersection of circumcircles of triangles $AEF$ and $ABC$. Prove that $EF$ bisects $A'K$. We define again point $K$ is reflection of $A'$ through $EF$ We have: $\angle{BEA'} = \angle{BAC} = \angle{CFA'}$ Then: $EA'$ $\parallel$ $AC$, $FA'$ $\parallel$ $AB$ or $EAFA'$ is parallelogram So: $\angle{EKF} = \angle{EA'F} = \angle{EAF}$ or $K$ $\in$ $(AEF)$ Then: $\angle{KEB} = 180^o - \angle{KEA} = 180^o - \angle{KFA} = \angle{KFC}$ But: $\dfrac{KE}{BE} = \dfrac{EA'}{BA'} = \dfrac{FA'}{CA'} = \dfrac{KF}{CF}$ so: $\triangle KBE$ $\sim$ $\triangle KCF$ Hence: $\angle{BKC} = \angle{BKE} + \angle{CKE} = \angle{CKF} + \angle{CKE} = \angle{EKF} = \angle{BAC}$ or $K$ $\in$ $(ABC)$