In the figure below, $E$ is the midpoint of the arc $ABEC$ and the segment $ED$ is perpendicular to the chord $BC$ at $D$. If the length of the chord $AB$ is $l_1$, and that of the segment $BD$ is $l_2$, determine the length of $DC$ in terms of $l_1, l_2$.
[asy]
unitsize(1 cm);
pair A=2dir(240),B=2dir(190),C=2dir(30),E=2dir(135),D=foot(E,B,C);
draw(circle((0,0),2)); draw(A--B--C); draw(E--D); draw(rightanglemark(C,D,E,8));
label("$A$",A,.5A); label("$B$",B,.5B); label("$C$",C,.5C); label("$E$",E,.5E); label("$D$",D,dir(-60));
[/asy]
1. **Identify the key points and properties:**
   - \( E \) is the midpoint of the arc \( ABEC \).
   - \( ED \) is perpendicular to the chord \( BC \) at \( D \).
   - \( AB = l_1 \) and \( BD = l_2 \).

2. **Introduce point \( P \):**
   - Extend \( CB \) and take a point \( P \) such that \( PD = DC \).

3. **Analyze the properties of the segments and angles:**
   - Since \( E \) is the midpoint of the arc \( ABEC \), \( E \) is equidistant from \( A \), \( B \), and \( C \). Therefore, \( EA = EB = EC \).
   - \( EP = EC \) because \( E \) is the midpoint of the arc and \( P \) is on the extension of \( CB \).

4. **Use the isosceles triangle properties:**
   - Since \( \triangle EPC \) is isosceles with \( EP = EC \), we have \( \angle EPC = \angle ECP \).
   - Similarly, since \( \triangle EPA \) is isosceles with \( EP = EA \), we have \( \angle EPA = \angle EAP \).

5. **Relate the angles:**
   - From the properties of the circle, \( \angle EAB = \angle ECB \) because they subtend the same arc \( EB \).
   - Using the isosceles triangle properties, we have:
     \[
     \angle BPA = \angle EPA - \angle EPB = \angle EAP - \angle ECP = \angle EAP - \angle EAB = \angle BAP
     \]
   - Therefore, \( \triangle BPA \) is isosceles, implying \( BP = AB \).

6. **Calculate the length \( DC \):**
   - Since \( DC = DP \) and \( DP = DB + BP \), we have:
     \[
     DC = DB + BP = l_2 + l_1
     \]

Conclusion:
\[
\boxed{l_2 + l_1}
\]