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Answer:
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 . . . . given
- ABNC 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 median
- MA = MN, AC = b, AB = c . . . . given
- ∠CMA ≅ ∠BMN . . . . vertical angles are congruent
- ΔCMA ≅ ΔBMN . . . . SAS congruence postulate
- AC ≅ NB . . . . CPCTC
- NB = b . . . . substitution property of equality/congruence
- ∠CMN ≅ ∠ BMA . . . . vertical angles are congruent
- ΔCMN ≅ ΔBMA . . . . SAS congruence postulate
- AB ≅ NC . . . . CPCTC
- NC = c . . . . substitution property of equality
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* 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.
