Name the postulate or theorem you can use to prove npm=omp

ΔNPM and ΔOMP by SAS postulate

B is correct

Step-by-step explanation:

In ΔNPM and ΔOMP

    NP = OM              (Given)

∠NPM = ∠OMP         (Given)

     PM = MP              ( Common )

So, ΔNPM ≅ ΔOMP by SAS property of concurrency.

ΔNPM and ΔOMP are congruent because their two side and angle between them are equal.

Therefore, SAS postulate use here

Hence, ΔNPM and ΔOMP by SAS postulate

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step-by-step explanation: vsfdvsfvd