The sides $AD$ and $BC$ of a convex quadrilateral $ABCD$ are extended to meet at $E$.  Let $H$ and $G$ be the midpoints of $BD$ and $AC$, respectively.  Find the ratio of the area of the triangle $EHG$ to that of the quadrilateral $ABCD$.
1. **Identify Key Points and Midpoints:**
   - Let \( H \) and \( G \) be the midpoints of \( BD \) and \( AC \), respectively.
   - Extend sides \( AD \) and \( BC \) to meet at point \( E \).
   - Let \( F \) be the intersection of \( AB \) and \( DC \).
   - Let \( K \) be the midpoint of \( EF \).

2. **Newton Line:**
   - The points \( G \), \( H \), and \( K \) are collinear on the Newton line of the convex quadrilateral \( ABCD \).

3. **Area of Boomerang \( EGFH \):**
   - The area of the "boomerang" \( EGFH \) is twice the area of \( \triangle EHG \).

4. **Height Considerations:**
   - Let \( h_P \) denote the distance from a point \( P \) to the line \( EF \).
   - Assume without loss of generality (WLOG) that \( H \) is between \( G \) and \( K \).

5. **Area Calculation:**
   - The area of \( [EGFH] \) can be expressed as:
     \[
     [EGFH] = [GEF] - [HEF] = \frac{1}{2} EF (h_G - h_H)
     \]
   - Since \( H \) and \( G \) are midpoints, we have:
     \[
     h_G = \frac{h_A + h_C}{2} \quad \text{and} \quad h_H = \frac{h_B + h_D}{2}
     \]
   - Therefore:
     \[
     h_G - h_H = \frac{h_A + h_C}{2} - \frac{h_B + h_D}{2} = \frac{h_A + h_C - h_B - h_D}{2}
     \]
   - Substituting back, we get:
     \[
     [EGFH] = \frac{1}{2} EF \left( \frac{h_A + h_C - h_B - h_D}{2} \right) = \frac{1}{4} EF (h_A + h_C - h_B - h_D)
     \]

6. **Relating to Quadrilateral \( ABCD \):**
   - The area of quadrilateral \( ABCD \) can be expressed as:
     \[
     [ABCD] = [CEF] + [AEF] - [BEF] - [DEF]
     \]
   - Since \( 2[EGFH] = [CEF] + [AEF] - [BEF] - [DEF] \), we have:
     \[
     2[EGFH] = [ABCD]
     \]
   - Therefore:
     \[
     2[EGFH] = 4[EHG] = [ABCD]
     \]

7. **Ratio of Areas:**
   - The ratio of the area of \( \triangle EHG \) to that of quadrilateral \( ABCD \) is:
     \[
     \frac{[EHG]}{[ABCD]} = \frac{1}{4}
     \]

The final answer is \(\boxed{\frac{1}{4}}\).