Because $MN$ is parallel to $BD$ and $MN=BD/2$ and $PQ$ is parralel to $BD$ and $PQ=BD/2$ we have that $MNPQ$ is parrallelogram. Now because $ABCD$ is cyclic we have $\angle MEN+\angle QEP=180$ and $MNPQ$ is parrallelogram $\implies k1=k2$ now if we take $O1,O2$ centers of $k1,k2$ we have $O1EO2F$ is a parallelogram which now $\implies$ $\angle FEN=\angle DBA$ and by angle chase we have $\angle AEN=90-\angle DBA$ so we are done.

Wolowizard wrote: Because $MN$ is parallel to $BD$ and $MN=BD/2$ and $PQ$ is parralel to $BD$ and $PQ=BD/2$ we have that $MNPQ$ is parrallelogram. Lemma: If $ABCD$ is cyclic and $S$ is midpoint of a side and $X$ is intersection of diagonals than $XS$ is perpendicular to opposite side. Proof: Trivial angle chase. The Lemma isn't right.

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My solution: Let $ O_1, O_2 $ be the center of $ \odot (EMN), \odot (EPQ) $, respectively . From $ \triangle EAB \cup N \sim \triangle EDC \cup Q \Longrightarrow \angle NEB=\angle QEC $ . ... (1) From $ \triangle EBC \cup P \sim \triangle EAD \cup M \Longrightarrow \angle MED=\angle PEC $ . ... (2) From (1) and (2) we get $ \angle NO_1M=\angle PO_2Q $ , so combine with $ PQ \parallel MN $, $ PQ=MN \Longrightarrow \triangle NO_1M $ and $ \triangle PO_2Q $ are congruent and homothety $ \Longrightarrow O_1O_2 \parallel AC $ . i.e. $ EF \perp AC $ Q.E.D

Almost exactly the same as problem 1 of British Mathematical Olympiad 2012 – Round 2: The diagonal AC and BD of a cyclic quadrilateral meet at E. The midpoints of the sides AB, BC, CD and DA are P, Q, R and S respectively. Prove that the circles EPS and EQR have the same radius. http://www.bmoc.maths.org/home/bmo2-2012.pdf

Yes there was word about that. I think the Authors knew that.But many solved it ($27$ out of $45 $had more than $6$ out of $7$ points)

Lemma: Let the circumcenter of $\Delta EMN $ is $ O_1$ and the circumcenter of $\Delta EPQ$ is $O_2$. Then $AC \parallel O_1O_2$ Proof: Let $O_1X_1 \perp MN$ and $O_2X_2 \perp PQ$. Since $(EMN) = (EPQ)$, we see that $O_1X_1 = O_2X_2$ and $MN \parallel PQ$, then it's pretty easy to see that $O_1X_1O_2X_2$ is actually a parallelogram. Then, $$O_1O_2 \parallel X_1X_2 \parallel MQ \parallel PN \parallel AC$$ If we use it in our problem, $AC \parallel O_1O_2 \perp EF$. So we are done.

Obviously $\angle ABE=\angle ABD=\angle ACD=\angle ECD$ and $\angle AEB=\angle CED$. So, triangles $AEB$ and $CED$ are similar. $\frac{AB}{CD}=\frac{BE}{CE}$ and $\frac{\frac{AB}{2}}{\frac{CD}{2}}=\frac{BE}{CE}$ so $\frac{BN}{CQ}=\frac{BE}{CE} \implies BEN$ and $CEQ$ are similar. Now we have lot of angle chase: $\angle MFE=\angle MNE=\angle BEN=\angle QEC=\angle MQE$ and $\angle QFE=\angle EMQ$ so point $F$ is orthocenter of triangle $EMQ$ and $EF \perp QM$. $QM$ is middleline of triangle $ACD$ so $QM \parallel AC$ and $EF \perp AC$

nguyenvanthien63 wrote: Wolowizard wrote: Because $MN$ is parallel to $BD$ and $MN=BD/2$ and $PQ$ is parralel to $BD$ and $PQ=BD/2$ we have that $MNPQ$ is parrallelogram. Lemma: If $ABCD$ is cyclic and $S$ is midpoint of a side and $X$ is intersection of diagonals than $XS$ is perpendicular to opposite side. Proof: Trivial angle chase. The Lemma isn't right. This lemma is true if and only if $AC\perp BD$. This is called ''Brahmagupta Theorem''.