Answer:
Given: a circle with centre O on which two tangents PM and PN are drawn from an external point P. To prove: PM = PN Construction: Join OM, ON and OP. Proof: Since tangent and radius are perpendicular at point of contact, \[\therefore \] \[\angle OMP=\angle ONP=90{}^\circ \] In \[\Delta \text{ }POM\]and \[\Delta \text{ }PON\], \[OM=ON\] (Radii) \[\angle OMP=\angle ONP\] \[PO=OP\] (Common) \[\therefore \] \[\Delta \,OMP\cong \Delta \,ONP\] (RHS cong.) \[\therefore \] \[PM=PN\] (CPCT) Hence Proved.
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