Question

in the given figure p is any point on the bisector of angle aob if pm perpendicular oa and PN perpendicular OB prove that pm equals to PN​

Answer 1

In ∆OMP and ∆ONP, ∠MOP = ∠NOP [because OP is bisector]

Side OP is common to both triangles

Side OM = ON (we assumed)

Hence, ∆OMP and ∆ONP are congruent

∴ Side PM = Side PN