See that $\stackrel{\frown}{PQ}=\stackrel{\frown}{PB}+\stackrel{\frown}{CQ}\implies\angle BAC=45^\circ$. Best regards, sunken rock

Let, $H$ be the orthocentre of $\triangle APQ$. $\therefore \angle A=180^{\circ}-\angle PHQ=\angle PCQ=90^{\circ}-\angle A \Rightarrow 2\angle A=90^{\circ}\Rightarrow \angle A=45^{\circ}$

Let$H$ be the $orthocentre$ As $\angle PHQ=180-$$\angle A$. As $quadrilateral BPHQ$is $cyclic$ so $\angle PBQ=A$. $\angle BQA=90$(angle subtended by diameter) So $\angle A=45$ Q.E.D

$J$ is the orthocenter of $\bigtriangleup APQ \implies \angle PJQ=180^{\circ}-\angle A$.Since,$J$ lies on $\Gamma,\angle PJQ=180^{\circ}-(90^{\circ}-\angle A) \implies \angle A=45^{\circ}$.

sunken rock wrote: See that $\stackrel{\frown}{PQ}=\stackrel{\frown}{PB}+\stackrel{\frown}{CQ}\implies\angle BAC=45^\circ$. Best regards, sunken rock How did you get these.....