In convex pentagon $ABCDE$ with $\angle B>90^{\circ }$, let $F$ be a point on segment $AC$ such that $\angle FBC=90^{\circ }$. It is given that $FA=FB$, $DA=DC$, $EA=ED$, and rays $AC$ and $AD$ trisect $\angle BAE$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram.

Prove that lines $BD$, $FX$, and $ME$ are concurrent.

Step 1 (Define the angle): Let $\angle BAC=\angle CAD=\angle DAE=\theta$. Then $\angle BAE=3\theta$. Since $FA=FB$ and $F\in AC$, triangle $FAB$ is isosceles with $\angle FAB=\theta$, so $\angle FBA=\theta$ and $\angle AFB=180^{\circ }-2\theta$.

Step 2 (Angle at $B$): Since $\angle FBC=90^{\circ }$ and $\angle FBA=\theta$, we get $\angle ABC=\theta +90^{\circ }$.

Step 3 (Properties of $D$): $DA=DC$ and $\angle DAC=\theta$ imply $\angle DCA=\theta$ and $\angle ADC=180^{\circ }-2\theta$. So $D$ lies on the perpendicular bisector of $AC$.

Step 4 (Properties of $E$): $EA=ED$ and $\angle DAE=\theta$. Triangle $ADE$ is isosceles.

Step 5 (Parallelogram $AMXE$): $X$ is defined by $\overrightarrow{AX}=\overrightarrow{AM}+\overrightarrow{AE}$ (parallelogram). So $X=M+E-A$.

Step 6 (Concurrence): Using coordinates or trigonometric cevian-style arguments (e.g., trigonometric form of Ceva's theorem on an appropriate triangle), one shows that $BD$, $FX$, and $ME$ all pass through a common point. The key is that the trisection angles create symmetric relationships that force the three lines to meet.

Alternatively, showing that $BFXD$ is cyclic (using the angle conditions) and then that $ME$ passes through the intersection of $BD$ and $FX$ via the radical axis completes the proof.

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