We have shown that MN∥BC in triangle ABC under the given conditions.
To prove that MN is parallel to BC in triangle ABC with BM=CN and ∠B=∠C, we can use the Converse of the Corresponding Angles Postulate.
Given:
△ABC
M on AB and N on AC such that BM=CN
∠B=∠C
To prove:
MN∥BC
Proof:
Extend BM and CN to meet at point O.
Since
∠B=∠C, ∠BOC is a straight angle (180∘).
In triangles BMO and CNO:
∠BMO=∠CNO (vertically opposite angles)
∠BMN=∠CON (corresponding angles as
BM∥CO and
BN∥CO)
BM=CN (given)
By Angle-Angle-Side (AAS) congruence, triangles BMO and CNO are congruent.
Therefore, BO=CO (corresponding parts of congruent triangles).
Since BM=CN and BO=CO, MO is the perpendicular bisector of BC.
By the Converse of the Perpendicular Bisector Theorem,
MN is parallel to BC.
Hence, we have shown that MN∥BC in triangle ABC under the given conditions.
Question
In a ΔABC, M and N are points on the sides AB and AC respectively such that BM = CN. If ∠B=∠C then show that MN || BC