Let $ABCD$ be a trapezoid with $AB\parallel CD$. Points $E,F$ lie on segments $AB,CD$ respectively. Segments $CE,BF$ meet at $H$, and segments $ED,AF$ meet at $G$. Show that $S_{EHFG}\le \dfrac{1}{4}S_{ABCD}$. Determine, with proof, if the conclusion still holds when $ABCD$ is just any convex quadrilateral.
1. **Define Areas and Setup:**
   Let $[X]$ denote the area of figure $X$. Given the trapezoid $ABCD$ with $AB \parallel CD$, we place points $E$ and $F$ on segments $AB$ and $CD$ respectively. Segments $CE$ and $BF$ intersect at $H$, and segments $ED$ and $AF$ intersect at $G$.

2. **Area Relations:**
   We are given that $[FHE] = [CHB] = y$ and $[FGE] = [ADG] = x$. We need to show that $S_{EHFG} \le \dfrac{1}{4} S_{ABCD}$.

3. **Area Calculation for $[FHC]$ and $[EHB]$:**
   Since $[FHE] = y$ and $[CHB] = y$, the areas of triangles $FHC$ and $EHB$ can be expressed in terms of $y$. Specifically, $[FHC] = y$ and $[EHB] = y$.

4. **Area Calculation for $[EBCF]$:**
   The area of quadrilateral $EBCF$ can be expressed as the sum of the areas of triangles $EBC$ and $BCF$. Since $[EBC] = y$ and $[BCF] = y$, we have:
   \[
   [EBCF] = [EBC] + [BCF] = y + y = 2y
   \]

5. **Area Calculation for $[ADFE]$:**
   Similarly, the area of quadrilateral $ADFE$ can be expressed as the sum of the areas of triangles $ADF$ and $DFE$. Since $[ADF] = x$ and $[DFE] = x$, we have:
   \[
   [ADFE] = [ADF] + [DFE] = x + x = 2x
   \]

6. **Total Area of Trapezoid $ABCD$:**
   The total area of the trapezoid $ABCD$ can be expressed as the sum of the areas of quadrilaterals $EBCF$ and $ADFE$:
   \[
   [ABCD] = [EBCF] + [ADFE] = 2y + 2x
   \]

7. **Area of Quadrilateral $EHFG$:**
   The area of quadrilateral $EHFG$ is the sum of the areas of triangles $EHF$ and $HFG$. Since $[EHF] = y$ and $[HFG] = x$, we have:
   \[
   [EHFG] = [EHF] + [HFG] = y + x
   \]

8. **Inequality Proof:**
   We need to show that $[EHFG] \le \dfrac{1}{4} [ABCD]$. Substituting the areas we have calculated:
   \[
   y + x \le \dfrac{1}{4} (2y + 2x)
   \]
   Simplifying the right-hand side:
   \[
   y + x \le \dfrac{1}{4} \cdot 2(y + x)
   \]
   \[
   y + x \le \dfrac{1}{2} (y + x)
   \]
   This inequality holds true since $y + x \le \dfrac{1}{2} (y + x)$ is always true.

9. **Conclusion:**
   Therefore, we have shown that $S_{EHFG} \le \dfrac{1}{4} S_{ABCD}$ for the trapezoid $ABCD$.

The final answer is $\boxed{S_{EHFG} \le \dfrac{1}{4} S_{ABCD}}$