Answer:
Given, a circle of radius OA and centred at O with chord AB and tangents PQ & RS are drawn from point A and B respectively. Draw \[OM\bot AB\], and join OA and OB. In \[\Delta \text{ }OAM\]and \[\Delta \text{ O}MB\], \[OA=OB\] (Radii) \[OM=OM\] (Common) \[\angle OMA=\angle OMB\] (Each \[90{}^\circ \]) \[\therefore \] \[\Delta \,OAM\cong \Delta \,OMB\] (R.H.S. cong.) \[\therefore \] \[\angle OAM=\angle OBM\] (CECT) Also, \[\angle OAP=\angle OBR=90{}^\circ \] (Line joining point of contact of tangent to centre is perpendicular on it) On addition, \[\angle OAM+\angle OAP=\angle OBM+\angle OBR\] \[\Rightarrow \] \[\angle PAB=\angle RBA\] \[\Rightarrow \] \[\angle PAQ-\angle PAB=\angle RBS-\angle RBA\] \[\Rightarrow \] \[\angle QAB=\angle SBA\] Hence Proved
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