jmc

We begin by drawing a diagram. Since $△ABC$ is isosceles with $AB=BC$, altitude $\overline{BE}$ is also a median: $E$ is the midpoint of $\overline{AC}$. Thus, $AE=EC=6/2=3$.



First we determine the area of $△ABC$. We determine $BE$, the height of the triangle, by using the Pythagorean Theorem on right triangle $△BAE$. This gives $$BE=\sqrt{A{B}^2-A{E}^2}=\sqrt{5^2-3^2}=4.$$Thus, $$[△ABC]=\frac{1}{2}(BE)(AC)=\frac{1}{2}(4)(6)=12.$$Notice that we can compute the area of triangle $ABC$ in another way: by using $\overline{BC}$ as the base (instead of $\overline{AC}$), and using $\overline{AD}$ as the altitude. We know that $BC=5$ and $[△ABC]=12$, so we have $$\frac{1}{2}(5)(AD)=12.$$Solving yields $AD=24/5$.

Now, we can compute $DC$ by using the Pythagorean Theorem on right triangle $△ADC$: $$DC=\sqrt{A{C}^2-A{D}^2}=\sqrt{6^2-(24/5{)}^2}=18/5.$$With this value, we can compute the area of triangle $ADC$: $$[△ADC]=\frac{1}{2}(AD)(DC)=\frac{1}{2}\left(\frac{24}{5}\right)\left(\frac{18}{5}\right)=\frac{216}{25}.$$Both triangle $DEA$ and triangle $DEC$ share the altitude from $D$ to $\overline{AC}$, and both triangles have equal base length. Thus, triangles $△DEA$ and $△DEC$ have the same area. Since $$[△DEA]+[△DEC]=[△ADC],$$we conclude $$[△DEC]=\frac{1}{2}\cdot \frac{216}{25}=\boxed{\frac{108}{25}}.$$