1. **Problem statement:** We have rectangle ABCD and a point F inside it such that $FA = FD$. We need to: - Place point F with justification. - Construct point E symmetric to F with respect to line $AB$. - Find the symmetric of segment $[AF]$ with respect to $AB$. - Show that $FD = AE$. 2. **Placing point F:** Since $FA = FD$ and both points A and D are vertices of the rectangle, point F lies on the perpendicular bisector of segment $AD$ inside the rectangle. Because ABCD is a rectangle, $AD$ is vertical or horizontal; thus, the perpendicular bisector is the line equidistant from A and D. So, F is on the line parallel to $AB$ passing through the midpoint of $AD$. 3. **Constructing point E symmetric to F with respect to $AB$:** The symmetry with respect to line $AB$ means reflecting F across $AB$. If $AB$ is horizontal, then E has the same horizontal coordinate as F but its vertical coordinate is symmetric about $AB$. If $F = (x_F,y_F)$ and $AB$ lies on $y = y_{AB}$, then $$E = (x_F, 2y_{AB} - y_F)$$ 4. **Symmetric of segment $[AF]$ with respect to $AB$:** The segment $[AF]$ connects points A and F. Reflecting $[AF]$ about $AB$ means reflecting both endpoints: - $A$ lies on $AB$, so its reflection is itself. - $F$ reflects to $E$. Thus, the symmetric segment is $[AE]$. 5. **Showing $FD = AE$:** Since $E$ is the reflection of $F$ about $AB$, and $D$ is the reflection of $A$ about $AB$ (because ABCD is a rectangle), triangles $AFD$ and $A E F$ are congruent by reflection properties. Therefore, $$FD = AE$$ This completes the proof. **Summary:** - $F$ lies on the perpendicular bisector of $AD$ inside the rectangle. - $E$ is the reflection of $F$ about $AB$. - The symmetric of $[AF]$ about $AB$ is $[AE]$. - By symmetry and congruence, $FD = AE$.