a) $Q$ is the center of the similarity $\rho$ mapping $B$ to $D$ and $E$ to $C$. Since $BE = CD$, its dilation factor is 1, i.e., it is a rotation. In particular, calling $Q_1$ and $Q_2$ the projections of $Q$ onto $(AB)$ and $(CD)$, since $\rho$ maps $Q_1$ to $Q_2$, $Q_1 = Q_2$, i.e., $Q$ lies on the bisector (the attentive reader will note that it needs to be verified that it is indeed the internal bisector, which can be easily done by a continuity argument by considering the extremal case).

b) We first observe that $\rho$ maps $K$ to $L$. In particular, $QK = QL$, and thus $RK = RL$. Moreover, the angle of the rotation is $\widehat{KQL}$ but is also, since the line $(EB)$ is mapped to the line $(CD)$, the angle between the lines $(BA)$ and $(CA)$. In particular, $A, K, Q$, and $L$ are concyclic by the inscribed angle theorem. Since it is also clear that $K, Q, L$, and $R$ are concyclic, $K, Q, L, A$, and $R$ are concyclic. The goal of the exercise is to show that $R$ is the center of the similarity mapping $K$ to $L$ and $B$ to $C$. Let $R'$ be the center of this similarity. $R'$, like $R$, lies on the circumcircle of $KQL$. Furthermore, since $KB = LC$, this similarity is a rotation, $d'$, where, as for $R$, $R'K = R'L$. Thus, $R$ and $R'$ are among the two points $d'$ intersection of the perpendicular bisector of $[KL]$ and the circumcircle of $KAL$, and a somewhat convoluted argument of positioning (the meticulous reader will note that it can be formalized without too much difficulty) shows that they are in fact the same. Hence the conclusion.