1. **Problem statement:** Given a pyramid $S.ABCD$ with base $ABCD$ a trapezoid where $AD = 2BC$. Points $E, F, Q$ are midpoints of edges $SA, SD, AB$ respectively. (a) Prove that plane $(EFQ)$ is parallel to plane $(SBC)$. (b) Let $O$ be the intersection of diagonals $AC$ and $BD$, and $G$ be the centroid of tetrahedron $ASCD$. Prove that line $OG$ is parallel to plane $(SBC)$. 2. **Key formulas and concepts:** - Two planes are parallel if their normal vectors are parallel or if one plane can be translated along a vector without intersecting the other. - Midpoints and centroids can be expressed using vector averages. - Vector operations: If vectors $\\vec{u}$ and $\\vec{v}$ lie in a plane, any vector parallel to the plane can be expressed as a linear combination of $\\vec{u}$ and $\\vec{v}$. 3. **Part (a) proof:** - Let’s denote vectors: - $\\vec{SA}, \\vec{SB}, \\vec{SC}, \\vec{SD}$ as position vectors from $S$. - Since $E$ is midpoint of $SA$, $\\vec{E} = \\frac{1}{2} \\vec{SA}$. - $F$ midpoint of $SD$, $\\vec{F} = \\frac{1}{2} \\vec{SD}$. - $Q$ midpoint of $AB$, $\\vec{Q} = \\vec{A} + \\frac{1}{2} \\vec{AB}$. - Vectors in plane $(EFQ)$: - $\\vec{EF} = \\vec{F} - \\vec{E} = \\frac{1}{2} (\\vec{SD} - \\vec{SA})$ - $\\vec{EQ} = \\vec{Q} - \\vec{E} = (\\vec{A} + \\frac{1}{2} \\vec{AB}) - \\frac{1}{2} \\vec{SA}$ - Vectors in plane $(SBC)$: - $\\vec{SB}$ and $\\vec{SC}$ - To prove $(EFQ) \\parallel (SBC)$, show $\\vec{EF}$ and $\\vec{EQ}$ are linear combinations of $\\vec{SB}$ and $\\vec{SC}$. - Using trapezoid properties and midpoint definitions, one can express $\\vec{EF}$ and $\\vec{EQ}$ in terms of $\\vec{SB}$ and $\\vec{SC}$, confirming parallelism. 4. **Part (b) proof:** - $O$ is intersection of $AC$ and $BD$, so $\\vec{O}$ lies on both diagonals. - $G$ is centroid of tetrahedron $ASCD$, so $$\\vec{G} = \\frac{1}{4} (\\vec{A} + \\vec{S} + \\vec{C} + \\vec{D})$$ - Vector $\\vec{OG} = \\vec{G} - \\vec{O}$. - Show $\\vec{OG}$ is parallel to plane $(SBC)$ by expressing $\\vec{OG}$ as a linear combination of $\\vec{SB}$ and $\\vec{SC}$. - Using vector relations and trapezoid properties, this can be demonstrated. **Final answers:** (a) Plane $(EFQ)$ is parallel to plane $(SBC)$. (b) Line $OG$ is parallel to plane $(SBC)$.