CBSE Class 10 Maths (Basic) 2025 · Paper 430(B)/S · Question 23

In right triangles $OAP$ and $OBP$, $OA=OB$ (radii), $OP$ is common and $\angle OAP=\angle OBP=90^{\circ }$ (radius $\perp$ tangent), so $△OAP\cong △OBP$ by RHS. Hence $\angle AOP=\angle BOP$ (CPCT): the tangents $PA$ and $PB$ subtend equal angles at the centre. Hence proved.