Let $AD$ meets $BC$ in point $K$. $\angle ABC = \angle ADC$ so by condition $\angle EDC = \angle ACB = \angle ADB$ so $DE$ and $DA$ are isogonal conjugate in triangle $BDC$ and $AE$ and $DA$ are isogonal conjugate in triangle $ABC$ . It is well-known that $(\frac{BD}{CD})^2 = \frac{BK}{CK} \frac{BE}{CE} = (\frac{BA}{CA})^2 $ so $BD*AC = CD*AB$ and $ABCD$ is harmonic. Now let tangents to $B$ and $C$ and $AD$ meets in point $F$. $ \angle BFC = \angle AFC + \angle BFA = 180 - \angle DAC - \angle DCA - \angle DCF + 180 - \angle BAD - \angle ABD - \angle DBF = 360 - 2\angle BAC - 180 = 180 - 2\angle BAC $ and $ \angle FBS + \angle SCF = \angle ABD - \angle CBD + \angle BAD + \angle ACD - \angle BCD + \angle DAC = 180 - \angle BAC + \angle BAC = 180 $ Here we use that BC is symedian in triangle $ABD$ and $CB$ is symedian in triangle $ACD$. So quadrilateral $BSCF$ is cyclic. And $\angle BSC = 180 - \angle BFC = 2\angle BAC$. Done.

Another (easier) solution???

Show $BD \cdot AC =CD \cdot AB (*) $ like @2above , we can apply Ptolemy's Theorem to get $BC\cdot AD = AB\cdot CD + AC \cdot BD$. Combine with $(*)$ to get $\frac{SD}{CD}=\frac{AD}{2\cdot CD}=\frac{BA}{BC}$. By cyclic quadrilateral, we have $\angle ABC= \angle SDC$. Then,$\triangle ABC \sim \triangle SDC$. Let $O$ be the circumcenter. $\angle OBC=90^{ \circ} -\angle BAC=\angle OSC.$ Therefore, $BSOC$ is cyclic. Thus, $ 2 \angle BAC= \angle BOC= \angle BSC$. $\blacksquare$

Let $O$ be the circumcenter of triangle $ABC$. And let $R \in AD, R\in (BOC)$. We'll prove that $R=S$ $\iff$ $AR=RD$ $\iff$ $OR \perp AD$. If $DE \cap (ABC)=K$ and $AE \cap (ABC)$, $AK \parallel BC \parallel DL$. $OE \perp BC$ $AD,OE,KL$ are concurrence at $N$. $OE\cap KA=Q, CT \cap KA=P$, $\angle PCN=90=\angle PQN \implies P,Q,C,N$ are cyclic. $\angle OBC=\angle BCQ=\angle CQA=\angle CNP= \angle CNO \implies N \in BOC $ $\implies B,R,O,C,N$ are cyclic. $\angle BRN=\angle BAC \implies OR \perp AD$.

After consider $E$ is Dumpty point from $C$ on $ADC$ the question is trivial