1. **Stating the problem:** Given a polygon with points and segments where $EP = FP$, a quadrilateral $ABCD$ with $AC = BD$ and angles $\angle ACD = \angle BDC$, and a trapezoid $ABCD$ with $AB \parallel DC$, $E$ midpoint of $AD$, $BE$ extended to $F$ on $CD$ such that $FE = BE$. We also have $BD = AF$ and $BD \parallel AF$. 2. **Understanding the properties:** - $EP = FP$ means point $P$ is equidistant from $E$ and $F$. - $AC = BD$ and $\angle ACD = \angle BDC$ imply congruent triangles or isosceles properties. - $AB \parallel DC$ defines trapezoid properties. - $E$ midpoint of $AD$ means $AE = ED$. - $FE = BE$ and $BD = AF$ with $BD \parallel AF$ suggest parallelogram or congruent segments. 3. **Using the midpoint and parallel lines:** Since $E$ is midpoint of $AD$, $AE = ED$. Since $AB \parallel DC$, trapezoid properties apply. 4. **Proving $BD = AF$ and $BD \parallel AF$:** Given $BD = AF$ and $BD \parallel AF$, quadrilateral $BDAF$ is a parallelogram. 5. **Using $FE = BE$ and $E$ midpoint:** Since $FE = BE$ and $E$ is midpoint, triangle $BEF$ is isosceles with $BE = FE$. 6. **Conclusion:** The given conditions establish relationships of equal lengths and parallelism that define congruent triangles and parallelograms within the figures. Final answer: The conditions confirm $BD = AF$ and $BD \parallel AF$ with $FE = BE$ and $E$ midpoint of $AD$ in the trapezoid with $AB \parallel DC$.