Solution:

Let the tangent to the circumcircle of $\triangle A B C$ at point $A$ intersect line $B C$ at $T$. Since $A B \leq A C$ we get that $B$ lies between $T$ and $C$. Since $\text{B A T= A C B}$ and $\text{A B T= 180 - A B C}$ we get $\text{E T A= B T A= A B C- A C B= A D E}$ which gives that $A, E, D, T$ are concyclic. Since
$$\text{T D B+ B C A= T D B+ B D A= T D A= A E T= A C B+ E A C}$$
this means $\text{T D B= E A C= D A B}$ which means that $T D$ is tangent to the circumcircle of $\triangle A B C$ at point $D$.

Using similar triangles $T A B$ and $T C A$ we get
$$\frac{A B}{A C}=\frac{T A}{T C}$$
Using similar triangles $T B D$ and $T D C$ we get
$$\frac{B D}{C D}=\frac{T D}{T C}$$
Using the fact that $T A=T D$ with (1) and (2) we get
$$\frac{A B}{A C}=\frac{B D}{C D}$$
Now since $\text{D A B= C A E}$ and $\text{B D A= E C A}$ we get that the triangles $D A B$ and $C A E$ are similar. Analogously, we get that triangles $C A D$ and $E A B$ are similar. These similarities give us
$$\frac{D B}{C E}=\frac{A B}{A E} \quad \text{ and } \quad \frac{C D}{E B}=\frac{C A}{E A}$$
which, when combined with (3) give us $B E=C E$ giving $E$ is the midpoint of side $B C$.

Using the fact that triangles $D A B$ and $C A E$ are similar with the fact that $E$ is the midpoint of $B C$ we get:
$$\frac{2 D S}{C A}=\frac{D A}{C A}=\frac{D B}{C E}=\frac{D B}{\frac{C B}{2}}=\frac{2 D B}{C B}$$
implying that
$$\frac{D S}{D B}=\frac{C A}{C B}$$
Since $\text{S D B= A D B= A C B}$ we get from (4) that the triangles $S D B$ and $A C B$ are similar, giving us $\text{B S D= B A C}$. Analogously we get $\triangle S D C$ and $\triangle A B C$ are similar we get $\text{C S D= C A B}$. Combining the last two equalities we get
$$\text{2 B A C= B A C+ C A B= C S D+ B S D= C S B}$$
This completes the proof.

Alternative solution (PSC).

Lemma 1. A point $P$ is such that $\text{P X Y= P Y Z}$ and $\text{P Z Y= P Y X}$. If $R$ is the midpoint of $X Z$ then $\text{X Y P= Z Y R}$.

Proof. We consider the case when $P$ is inside the triangle $X Y Z$ (the other case is treated in similar way). Let $Q$ be the conjugate of $P$ in $\triangle X Y Z$ and let $Y Q$ intersects $X Z$ at $S$.

Then $\text{Q X Z= Q Y X}$ and $\text{Q Z X= Q Y Z}$ and therefore $\triangle S X Y \sim \triangle S Q X$ and $\triangle S Z Y \sim \triangle S Q Z$. Thus $S X^{2}=S Q \cdot S Y=S Z^{2}$ and we conclude that $S \equiv R$. This completes the proof of the Lemma.

For $\triangle D C A$ we have $\text{C D E= E C A}$ and $\text{E A C= E C D}$. By the Lemma 1 for $\triangle D C A$ and point $E$ we have that $\text{S C A= D C E}$. Therefore
$$\text{D S C= S A C+ S C A= S A C+ D C E= S A C+ B A D= B A C .}$$
By analogy, Lemma 1 applied for $\triangle B D A$ and point $E$ gives $\text{B S D= B A C}$. Thus, $\text{B S C=2 B A C}$.