India_2017

Join $DE$ and $PQ$. Let $S$ be the centre of the circle $\Gamma$ passing through $E,P,Q,D$. Let $P^{\prime }$ denote the reflection of $P$ in $BI$. Since $\angle PBI=\angle QBI$, it follows that $P^{\prime }$ lies on $BQ$. Since $E$ and $D$ are symmetric about the line $EI$, and $P$ and $P^{\prime }$ are also symmetric about $BI$, it follows that $P^{\prime }$ lies on $\Gamma$. Since $Q$ lies on $\Gamma$ and $P^{\prime }$ lies on the line $BQ$, we conclude that $Q=P^{\prime }$. 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 $△AIC$ 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 $BA=BC$.