This one can also be done with the help of Alternate segment theorem

Let AR intersect smaller circle at N

Let AQ intersect smaller circle at M

Join MN

Draw a tangent a A

Let Y be the point on tangent such that angle RAY is acute

Let Z be the point on tangent such that angle RAY is obtuse

The following relation follows from Alternate Segment theorem

RAY=RQA

PAY=PMA

QAZ=QRA

PAZ=PNA

PAQ=PAZ-QAZ

=PNA-QRA

=RPN

=PAR(Alternate ST,smaller circle, tangent at P, chord PN)

Hence, proved.