Please help to solve this question, and provide the formal proof. Very much appreciated, and have a great day!

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  NB = b

  NC = c

Step-by-step explanation:

A "quick and dirty" proof could be this:

M is the midpoint of AN; M is the midpoint of BC . . . . givenABNC is a parallelogram . . . . diagonals AN and BC have the same midpoint (M)*NB = AC = b; NC = AB = c . . . . opposite sides of a parallelogram are congruent.

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More formally, we might prove it like this:

MC = MB . . . . definition of medianMA = MN, AC = b, AB = c . . . . given∠CMA ≅ ∠BMN . . . . vertical angles are congruentΔCMA ≅ ΔBMN . . . . SAS congruence postulateAC ≅ NB . . . . CPCTCNB = b . . . . substitution property of equality/congruence∠CMN ≅ ∠ BMA . . . . vertical angles are congruentΔCMN ≅ ΔBMA . . . . SAS congruence postulateAB ≅ NC . . . . CPCTCNC = c . . . . substitution property of equality

* The proof is "dirty" only in that the theorem "if the midpoints of the diagonals of a quadrilateral are coincident, then that quadrilateral is a parallelogram" is not one usually available to cite in proofs. The "more formal" proof here essentially proves that the quadrilateral is a parallelogram.

Siciuenckwk the answer