Let ABCD be a square with side length $a$. Let $M$ be a point on $BC$. The line $AM$ meets the diagonal $BD$ at $K$. From $K$ draw the perpendicular $KL$ to $CD$, with foot $L$. Let $P$ be the reflection of $C$ with respect to $L$. The line $PK$ meets $AB$ at $F$. Prove that the quadrilateral $AFMP$ is an isosceles trapezoid.
If $\frac{BM}{MC}=n$, prove that $FB=\frac{na}{(2n+1)(n+1)}$.
While experimenting with ratios in a square ABCD of side length $a$, I first considered the case where $M$ is the midpoint of $BC$. From a carefully constructed diagram I conjectured that the quadrilateral $AFMP$ should be an isosceles trapezoid, which I then proved synthetically, obtaining $FB=\frac{a}{6}$. This led me to consider a generalization where $M$ is an arbitrary point on $BC$ such that $\frac{BM}{MC}=n$. Using a similar approach, I found that $AFMP$ is again an isosceles trapezoid and that $FB=\frac{na}{(2n+1)(n+1)}$.
I present a synthetic proof below and would be interested in alternative approaches (coordinates, vectors, or transformations), as well as possible generalizations.
Extend $KC$ to meet $AB$ at $T$ and let $KL$ meet $AB$ at $N$. We have $AD=DC$, $DK$ common and $\angle ADK=\angle KDC=90^\circ \Rightarrow \triangle AKD \cong \triangle DKC$ (SAS) $\Rightarrow AK=KC$. Since $L$ is the midpoint of $CP$ and $KL \perp CD$, it follows $KC=KP \Rightarrow AK=KP$ (1).
Also $\angle ABM=\angle CBT=90^\circ$, $AB=BC$, and $\angle BAM=\angle TCB$ (supplementary to $\angle DAK=\angle KCD$) $\Rightarrow \triangle ABM \cong \triangle TBC$ (ASA) $\Rightarrow BT=BM$. Further $\triangle TBK \cong \triangle MBK$ (SAS) $\Rightarrow KT=KM$. Moreover $\triangle TKF \sim \triangle KPC$ (isosceles) $\Rightarrow KT=KF \Rightarrow KM=KF$ (2).
Now $\angle MKF=\angle AKP$ (vertical) and from (1),(2) we get $\triangle AKP \sim \triangle FKM \Rightarrow \angle AMF=\angle MAP \Rightarrow FM \parallel AP$ (3), and $AK+KM=KP+KF \Rightarrow AM=FP$ (4). From (3),(4) it follows that $AFMP$ is an isosceles trapezoid.
For the second part, if $\frac{BM}{MC}=n$, then $\frac{BM}{BC}=\frac{n}{n+1} \Rightarrow BM=\frac{na}{n+1}$ (5). Draw $FZ \perp AB,CD$ with feet $F,Z$. Since $KL \parallel FZ$, by Thales we get $\frac{LZ}{PL}=\frac{KF}{KP}=\frac{KM}{AK}$. Since $BM \parallel AD$, it follows $\frac{KM}{KA}=\frac{BM}{AD}=\frac{BM}{BC}=\frac{n}{n+1}$,
hence $\frac{LZ}{PL}=\frac{TN}{NB}=\frac{n}{n+1} \Rightarrow \frac{NB}{TN}=\frac{n+1}{n} \Rightarrow \frac{NB}{TB}=\frac{NB}{BM}=\frac{n+1}{2n+1} \Rightarrow NB=\frac{na}{2n+1}$ (6).
Since $TN=NF$, we have $NB=\frac{FB+TB}{2}$=$\frac{FB+BM}{2}$, and substituting from (5),(6) we obtain $FB=\frac{na}{(2n+1)(n+1)}$.
