The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ with $S_{ABC} = S_{ADC}$ intersect at $E$. The lines through $E$ parallel to $AD$, $DC$, $CB$, $BA$
meet $AB$, $BC$, $CD$, $DA$ at $K$, $L$, $M$, $N$, respectively. Compute the ratio $\frac{S_{KLMN}}{S_{ABC}}$
1. Given a convex quadrilateral \(ABCD\) with diagonals \(AC\) and \(BD\) intersecting at \(E\), and the areas of triangles \(ABC\) and \(ADC\) are equal, i.e., \(S_{ABC} = S_{ADC}\).
2. Lines through \(E\) parallel to \(AD\), \(DC\), \(CB\), and \(BA\) meet \(AB\), \(BC\), \(CD\), and \(DA\) at points \(K\), \(L\), \(M\), and \(N\) respectively.
3. We need to compute the ratio \(\frac{S_{KLMN}}{S_{ABC}}\).

First, we note that since \(AKEN\) and \(ELCM\) are parallelograms, the following properties hold:
- \(AK = EN\) and \(KE = AN\)
- \(EL = CM\) and \(LE = EM\)

4. Since \(S_{ABC} = S_{ADC}\), the quadrilateral \(ABCD\) is divided into two equal areas by the diagonal \(AC\). Therefore, the total area of \(ABCD\) is \(2S_{ABC}\).

5. We need to show that the area of quadrilateral \(KLMN\) is half the area of \(ABCD\). To do this, we will use the properties of parallelograms and the fact that the lines through \(E\) are parallel to the sides of \(ABCD\).

6. Consider the parallelograms \(AKEN\) and \(ELCM\):
   - Since \(AKEN\) is a parallelogram, \([ANE] = [AKE] = [KNE]\).
   - Since \(ELCM\) is a parallelogram, \([EMC] = [ECL] = [LEM]\).

7. We know that \([ACD] = [ABC]\), which implies \([DNEM] = [KBLE]\).

8. Since \(NE \parallel AB\), we have \(\frac{AN}{AD} = \frac{BE}{BD}\).

9. Since \(KE \parallel AD\), we have \(\frac{BK}{AB} = \frac{BE}{BD}\).

10. Therefore, \([NEM] = [KBL]\) and \([KEL] = [DEM]\).

11. This implies that the area of quadrilateral \(KLMN\) is half the area of \(ABCD\), i.e., \([KLMN] = \frac{[ABCD]}{2}\).

12. Since \([ABCD] = 2S_{ABC}\), we have:
\[
[KLMN] = \frac{2S_{ABC}}{2} = S_{ABC}
\]

13. Therefore, the ratio \(\frac{S_{KLMN}}{S_{ABC}}\) is:
\[
\frac{S_{KLMN}}{S_{ABC}} = \frac{S_{ABC}}{S_{ABC}} = 1
\]

The final answer is \(\boxed{1}\)