Let $M$ be the midpoint of the arc $BC$ of $\odot(ABC).$ $\angle AQP=\angle ADP=\angle ACM$ $\Longrightarrow$ $M,Q,P$ are collinear. Since $MB^2=MC^2=MI^2=MQ \cdot MP$ $\Longrightarrow$ $MI$ is tangent of $\odot(PIQ)$ $\Longrightarrow$ $\angle IQM=\angle PIM=\angle IDP=\angle ACM.$ Thus if $QI$ cuts $\odot(ABC)$ again at $E,$ then $AE \parallel BC$ $\Longrightarrow$ $E$ is reflection of $A$ across the perpendicular bisector of $BC.$ Hence, by obvious symmetry, $J \in EM$ and $AEJI$ is isosceles trapezoid $\Longrightarrow$ $\angle MAQ=\angle MEQ=\angle MAJ$ $\Longrightarrow$ $\angle BAQ=\angle CAJ.$

Let $AI\cap \odot (\triangle ABC)=M$. $IJ\parallel BC\Longrightarrow \angle IDP=\angle MIJ\Longrightarrow \triangle MIJ\sim \triangle PID\Longrightarrow \frac{IJ}{ID}=\frac{MI}{PD}\Longrightarrow IJ.PD=MI.ID=AI.MD\Longrightarrow \frac{AI}{IJ}=\frac{PD}{DM} \bigstar$. ($\because \frac{AI}{ID}=\frac{MI}{MD}=\frac{b+c}{a}$) $\angle BPQ=\angle MAQ=\frac{\angle A-\angle CAQ}{2}\Longrightarrow P,Q,M$ are collinear. From $\angle IDP=\angle DIJ$ and $\bigstar\Longrightarrow \triangle IAJ\sim \triangle DPM\Longrightarrow \angle MAJ=\angle BPM=\angle QAM$. DONE