Hello, dodoldotkom!
$\displaystyle AC\text{ is a diameter.}$
$\displaystyle \text{Hence: }\:\text{arc}(ABC) \,=\,\text{arc}(ADC) \,=\,180^o$
$\displaystyle \angle DBC = 30^o \quad\Rightarrow\quad \text{arc}(DC) = 60^o$
. . $\displaystyle \text{arc}(AD) = 120^o \quad\Rightarrow\quad \angle ACD = 60^o$
$\displaystyle \text{In }\Delta ECD: \:\angle ECD = 60^o,\;\angle EDC = 35^o \quad\Rightarrow\quad \angle DEC = 85^o$
$\displaystyle \text{Therefore: }\:\angle AED = 95^o$