Join $DE$ and $PQ$. Let $S$ be the centre of the circle $\Gamma$ passing through $E, P, Q, D$. Let $P'$ denote the reflection of $P$ in $BI$. Since $\angle PBI = \angle QBI$, it follows that $P'$ lies on $BQ$. Since $E$ and $D$ are symmetric about the line $EI$, and $P$ and $P'$ are also symmetric about $BI$, it follows that $P'$ lies on $\Gamma$. Since $Q$ lies on $\Gamma$ and $P'$ lies on the line $BQ$, we conclude that $Q = P'$. This means $BI$ is the perpendicular bisector of $PQ$. Since $BI$ is already the perpendicular bisector of $ED$, it follows that $ED$ is parallel to $PQ$.
Consider the circumcircle of the triangle $AIC$. We have

$$\angle BIC = 180^\{\circ\} - (B/2 + C/2) = 90^\{\circ\} + (A/2).$$
Since $\angle BID = 90^\{\circ\}$, we see that $\angle DIC = A/2 = \angle IAC$. Hence $ED$ is tangent to the circumcircle of $\triangle AIC$ at $I$. Hence the circumcentre of the triangle $AIC$ lies on the line $BS$. This implies that $BS$ is the perpendicular bisector of $AC$ as well. It follows that $ED \parallel AC$. Thus $PQ \parallel ED \parallel AB$.
Since $BS$ is the perpendicular bisector of $PQ$ and that of $AC$, we see that $PQCA$ is an isosceles trapezium. Hence $P, Q, C, A$ are concyclic. This gives $\angle PAC = \angle QCA$. Therefore $\angle A = \angle C$ and hence $BA = BC$.