A circle is tangential to sides $AB$ and $AD$ of convex quadrilateral $ABCD$ at $G$ and $H$ respectively, and cuts diagonal $AC$ at $E$ and $F$.  What are the necessary and sufficient conditions such that there exists another circle which passes through $E$ and $F$, and is tangential to $DA$ and $DC$ extended?
1. **Correcting the Typo and Setting Up the Problem:**
   The problem should state that a circle is tangential to sides \(AB\) and \(AD\) of convex quadrilateral \(ABCD\) at \(G\) and \(H\) respectively, and cuts diagonal \(AC\) at \(E\) and \(F\). We need to find the necessary and sufficient conditions for the existence of another circle passing through \(E\) and \(F\) and tangential to \(DA\) and \(DC\) extended.

2. **Identifying the First Circle:**
   Let the first circle \((P)\) be tangential to sides \(AB\) and \(AD\) at points \(G\) and \(H\) respectively. This circle intersects the diagonal \(AC\) at points \(E\) and \(F\).

3. **Conditions for the Second Circle:**
   We need to find the conditions under which there exists another circle \((Q)\) that passes through \(E\) and \(F\) and is tangential to the extensions of \(DA\) and \(DC\).

4. **Center of the Second Circle:**
   The second circle \((Q)\) must be centered on the internal bisector of the angle \(\angle CDA\) and also on the perpendicular bisector of the segment \(EF\). Let \(Q\) be the intersection of the internal bisector of \(\angle CDA\) with the perpendicular bisector of \(EF\).

5. **Radical Axis and Tangent Lengths:**
   The diagonal \(AC \equiv EF\) is the radical axis of the circles \((P)\) and \((Q)\). For the circle \((Q)\) to be tangential to the lines \(DA\) and \(DC\) at points \(K\) and \(L\), the tangent lengths from the vertices \(A\) and \(C\) to the circles \((P)\) and \((Q)\) must be equal. Specifically, the tangent lengths \(AG = AK\) and \(CH = CL\) must hold.

6. **Equality of Tangent Lengths:**
   Since the vertices \(A\) and \(C\) lie on the radical axis of the circles \((P)\) and \((Q)\), the tangent lengths from these vertices to the circles must be equal:
   \[
   AG = AK \quad \text{and} \quad CH = CL
   \]
   Similarly, the tangent lengths from the vertices \(B\) and \(D\) to the circles must also be equal:
   \[
   BG = BH \quad \text{and} \quad DK = DL
   \]

7. **Summing the Tangent Lengths:**
   Summing the tangent lengths, we get:
   \[
   AB + CD = (AG + BG) + (CL + DL) = AK + BH + CH + DK
   \]
   \[
   = (BH + CH) + (DK + AK) = BC + DA
   \]

8. **Conclusion:**
   The necessary and sufficient condition for the existence of another circle passing through \(E\) and \(F\) and tangential to \(DA\) and \(DC\) extended is that the quadrilateral \(ABCD\) must be tangential. This means that the sum of the lengths of opposite sides must be equal:
   \[
   AB + CD = BC + DA
   \]

The final answer is \( \boxed{ AB + CD = BC + DA } \)