kamatadu wrote: Let $ABCD$ be a cyclic quadrilateral. Points $E$ and $F$ lie on the sides $AD$ and $CD$ in such a way that $AE = BC$ and $AB = CF$. Let $M$ be the midpoint of $EF$. Prove that $\angle AMC = 90^{\circ}$. Let point $N$ be the reflection of $A$ with respect to $M$. The quadrilateral $AENF$ will be a parallelogram, then the triangle $CFN$ will be congruent to the triangle $ABC$, from here we get that $AC=CN$. Therefore, $ACN$ is an isoscele triangle, and since $CM$ is the median, it will also be the height.

Interestingly, this is (essentially) identical with Problem 13 from Baltic Way 2022.

Parallel from $B$ intersects $(ABC)$ at $B'$. Notice that $B'CF$ and $B'AE$ are isoscele triangles. Let $P$ and $Q$ be the midpoints of $B'F$ and $B'E$. $\triangle B'PC$ is similar to $\triangle AQB'$ and take $M'$ such that these triangles are similar to $\triangle AM'C$ There is a spiral similarity with center $C$ that sends $B'P$ to $AM'$. Thus there exists a spiral similarity with center $C$ that sends $M'P$ to $AB'$ $\angle (M'P,AB') = \angle PCB'= \angle (AB',BQ)$ hence $M'P\parallel B'Q$. Similarly $M'Q\parallel B'P$ Hence $BPM'Q$ is rhomboid $M=M'$,$\angle AMC=90$

I like this! Let $G$ be the point such that $ABCG$ is a parallelogram. A bit of angle chasing (computing angles at $G$ combined with the fact that $\triangle GAE$ and $\triangle GCF$ are isosceles) yields $\angle EGF = 90^\circ$. This means $M$ is the circumcenter of $\triangle EGF$ and the perpendicular bisectors of $\overline{EG}$ and $\overline{GF}$ are perpendicular. But these bisectors are precisely $AM$ and $CM$, respectively. The end.