1. **Problem statement:** Given a pyramid $S.ABCD$ with base $ABCD$ a trapezoid where $AB \parallel CD$. Point $E$ lies on segment $SA$. Plane $(P)$ passes through $E$ and is parallel to lines $AB$ and $AD$. We need to find the intersections of plane $(P)$ with the lateral faces of the pyramid and identify the shape formed by these intersections. 2. **Key concepts:** - A plane parallel to two intersecting lines (here $AB$ and $AD$) is parallel to the plane containing those lines (the base plane $ABCD$). - The intersection of plane $(P)$ with each lateral face (triangles $SAB$, $SBC$, $SCD$, $SDA$) will be a line segment. - Since $(P)$ is parallel to $AB$ and $AD$, it is parallel to the base plane $ABCD$. 3. **Step-by-step solution:** - Since $E$ lies on $SA$, and $(P)$ is parallel to $AB$ and $AD$, $(P)$ is parallel to the base plane $ABCD$. - The lateral faces are triangles $SAB$, $SBC$, $SCD$, and $SDA$. - Intersection of $(P)$ with face $SAB$: - Since $(P)$ passes through $E$ on $SA$ and is parallel to $AB$, the intersection line in $SAB$ is through $E$ and parallel to $AB$. - Intersection of $(P)$ with face $SBC$: - $(P)$ is parallel to $AB$ and $AD$, so it is parallel to the base plane. - The intersection with $SBC$ is a line parallel to $BC$ (since $BC$ lies in the base plane) passing through a point on $SB$ determined by the parallelism and position of $(P)$. - Intersection of $(P)$ with face $SCD$: - Similarly, the intersection is a line parallel to $CD$ passing through a point on $SC$. - Intersection of $(P)$ with face $SDA$: - $(P)$ passes through $E$ on $SA$ and is parallel to $AD$, so the intersection line is through $E$ and parallel to $AD$. 4. **Shape formed by the intersections:** - The four intersection lines form a quadrilateral. - Since $(P)$ is parallel to the base plane $ABCD$, the intersection shape is a trapezoid similar to the base $ABCD$. **Final answer:** The intersections of plane $(P)$ with the lateral faces are four line segments parallel respectively to $AB$, $BC$, $CD$, and $DA$, forming a trapezoid similar to the base $ABCD$.