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If in triangle ABC m

User Onestone
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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

User Sergey Aldoukhov
by
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