From the condition $BD/DC = AB^2/AC^2$ we know that $AD$ is the $A$-symmedian of the triangle $ABC$. Let $a, b, c$ be the side lengths of the triangle, then we get $AF/FB = CD/DB = b^2/c^2$ hence $AF = b^2c/(b^2 + c^2)$ and similarly $AE = c^2b/(b^2 + c^2)$. Using these we get $AE \cdot AC = AF \cdot AB$ hence $B, E, C, F$ are concyclic. Let the circumcircle of the triangle $ABC$ be $\Gamma$ and let the circumcircle of the triangle $AEF$ be $\omega$. Let $AD$ meet $\Gamma$ for the second time at $Q$ and let $AS \cap BC = P$. Let the intersection of $\Gamma$ and $\omega$ other than $A$ be $M$. The radical axis of the circles $\Gamma$, $\omega$ and $(BCEF)$ are concurrent, let this point be $AM \cap EF \cap BC = L$. Let the line passing through $A$ and parallel to $BC$ intersect $\Gamma$ for the second time at $A'$. Since $\angle C = \angle AFE = \angle EAA'$ we find that $AA'$ and $\omega$ are tangent to each other.

Let us consider the spiral similarity $\sigma$ centred at $M$ and satisfying $F \to B, E \to C$. It is in fact the spiral similarity that sends the circle $\omega$ to the circle $\Gamma$, thus $\sigma$ also satisfies $S \to A, A \to A'$ and $R \to Q$. Since $AQ$ is a symmedian, the tangent lines to $\Gamma$ passing through $Q$ and $A$ intersect on the line $BC$. Therefore, under $\sigma$, the tangent lines to $\omega$ passing through $R$ and $S$ intersect on the line $EF$. In addition, simple angle chasing gives that $\angle BDF = \angle C$ and $\angle AEF = \angle B, \angle DEC = \angle A$ which means $\angle FED = \angle C$ hence, circumcircle of the triangle $DEF$ and the line $BC$ are tangent to each other. From here we find $LD^2 = LE \cdot LF = LB \cdot LC$, also since we know $(P,D;B,C) = -1$ we obtain that $L$ is the midpoint of the segment $PD$. We found that the lines $AA'$ and $BC$ are parallel, therefore we have $(AP;AD;AL, AA') = -1$ and projecting these lines to $\omega$ we find that $(S,R;M,A) = -1$, which means the tangent lines to $\omega$ passing through $S, R$ intersect on the line $AM$. In this case, this intersection point was also on the line $EF$, it must be $L$.

So, we find that the polar line of $L$ with respect to $\omega$ is $SR$, and since $AM \cap EF = L$, from Brokard's Theorem in quadrilateral $AMEF$, the lines $AF$ and $EM$ intersect on the polar line $SR$. Hence $AB$, $EM$, $SR$ are concurrent, and we also have $(T,L;F,E) = -1$. Furthermore, on the circle $\omega$ we have $(S,Q;A,M) = -1$ and under $\sigma$ its image is $(A,Q;M,A') = 1$, hence the points $A', M, P$ are collinear. Let $A'M \cap AQ = V$, then we obtain $(P,V;M,A') = -1$. Finally, by definition we have $AA' \parallel BC$ and under $\sigma$ it becomes

Now, we will work on the problem statement to finish the solution. We will prove that both conditions are equivalent to $\angle B = 2\angle C$ or $180^\circ + \angle B = 2\angle C$. Solution to both cases are identical, thus we only study the case where $\angle B > \angle C$. First, we know that $\angle MA' = \angle B - \angle C$ and under $\sigma$ we have $\angle SMA = \angle B - \angle C$. Also, we have $\angle AME = \angle AFE = \angle C$. $\angle B = 2\angle C$ is equivalent to the following statements:
$$(1)\ \angle B = 2\angle C \Leftrightarrow \angle SMA + \angle AMB = \angle B - \angle C + (180^\circ - \angle C) = 180^\circ \Leftrightarrow B, M, W, S \text{ are collinear.}$$
$$(2)\ \angle B = 2\angle C \Leftrightarrow \angle MA' = \angle B - \angle C = \angle C = \angle AME \Leftrightarrow P, M, E, A' \text{ are collinear.}$$
From the first condition, $BS$ bisecting $[TL]$ is equivalent to $\angle B = 2\angle C$. From the concurrency $AB, SR, EM$ we find that $SR$ bisecting $[AB]$ is equivalent to $EM$ bisecting $[AB]$. $EM$ bisecting $[AB]$ is equivalent to the condition $(EA, EB; EM, ED) = -1$, and reflecting these lines to the line $BC$ shows that it is in fact equivalent to $E, M, P$ being collinear. From the second condition this is equivalent to $\angle B = 2\angle C$ hence the required statement is proven.