Question

In the adjoining figure, AC > AB and AM is the bisector of ∠BAC. Prove that ∠AMC > ∠AMB.

Answer 1

Given:

• In triangle ABC, AC > AB.
• AM is the internal bisector of ∠BAC, meeting BC at M.

To Prove:

• ∠AMC > ∠AMB.

Proof (Step-wise):

1. Since AM bisects ∠BAC, we have ∠BAM = ∠MAC.   ...(Given)

2. Write the angle-sum relations in triangles AMB and AMC:

In ΔAMB: ∠AMB + ∠ABM + ∠BAM = 180°.

In ΔAMC: ∠AMC + ∠ACM + ∠CAM = 180°.

3. Subtract the two equalities and use ∠BAM = ∠CAM to get ∠AMC – ∠AMB = ∠ABM – ∠ACM.

So, ∠AMC > ∠AMB ⇔ ∠ABM > ∠ACM.

4. Because B, M, C are collinear with M between B and C, the rays BM and BC coincide and the rays CM and CB coincide.

Hence, ∠ABM = ∠ABC and ∠ACM = ∠ACB.

5. Therefore, ∠ABM > ∠ACM is equivalent to ∠ABC > ∠ACB.

6. In ΔABC, the larger side subtends the larger opposite angle.

Since AC > AB, the angle opposite AC (which is ∠ABC) is larger than the angle opposite AB (which is ∠ACB).

Property of triangles: Larger side ↔ Larger opposite angle.

7. Combining steps 3–6, we obtain ∠AMC > ∠AMB.

∠AMC > ∠AMB, as required.