imc

$△ABC$ is isosceles, hence $\angle ACB=\angle CAB$. The sum of internal angles of $△ABC$ can now be expressed as $\angle B+\frac{5}{2}\angle B+\frac{5}{2}\angle B=6\angle B$, hence $\angle B=30^{\circ }$, and each of the other two angles is $75^{\circ }$. Now we know that $\angle DCE=\angle ACB=75^{\circ }$. Finally, $△CDE$ is isosceles, hence each of the two remaining angles ($\angle D$ and $\angle E$) is equal to $\frac{180^{\circ }-75^{\circ }}{2}=\frac{105^{\circ }}{2}=\boxed{52.5}$.