P is a point outside the circle as shown. MP and NP are the tangents are drawn to the circle from this point. Then quadrilateral OMPN is symmetrical with respect to OP.
True
In △OMP and △ONP,
OM = ON [radius]
OP = OP [common side]
∠OMP = ∠ONP [right angle triangle]
So, by R.H.S criterion,
△OMP ≅ △ONP
That is the triangles on either side of OP are congruent.
Hence OMPN is symmetrical with respect to OP.