Consider $F$ to be the reflection of $E$ with respect to $DC$, $BF\cap DC=\{G\}$ and $FE\cap AB=\{K'\}$. It is easy to see that $DEF$ is equilateral and that triangles $EFB$ with $EDC$ are congruent, which means, by say spiral-similarity, that the quadrilaterals $DFGE$ and $CGEB$ are cyclic. Now, \[\angle ECG=15^{\circ}\Rightarrow \angle EDG=30^{\circ}\Rightarrow \angle EFG=30^{\circ}\]But \[\angle EGD=\angle CGB=\angle CEB=60^{\circ}\]which means that $FE\perp DC$, or $FE\perp AB$, which can be written as $FK'\perp AB$. Anyway, $\angle ADF=\angle ADC+\angle FDC=105^{\circ}+30^{\circ}=135^{\circ}$. At the same time, $F$ is the reflection of $E$ wrt $DC$, which leads to $\angle DFC=\angle DEC=135^{\circ}=\angle ADF$. But we also have $FC=EC=CB=AD$, which means that triangles $ADF$ and $CFD$ are congruent $\Rightarrow AF=FB$. Finally, $FK'\perp AB$, which means that $K=K'$, so $EK\perp AB$ and $AE=EB$. On the other hand, $AE=EB=CB=AD$ and $FE=FD$, hence triangles $ADF$ and $AEF$ are congruent, which leads to $\angle AED=\angle ADE=\angle ADC-\angle EDC=105^{\circ}-30^{\circ}=75^{\circ}\Rightarrow \angle AEK=180^{\circ}-\angle DEF-\angle AED=180^{\circ}-60^{\circ}-75^{\circ}=45^{\circ}$ Thus $AEB$ is a right angled isosceles triangle. It immediately follows that $\angle BKC=45^{\circ}$.