Question

In the adjoining figure ∠DBC > ∠BCE. Prove that AB > AC.

Answer 1

Given: In the adjoining figure BD is the extension of BA and CE is the extension of CA and ∠DBC > ∠BCE.

To Prove: AB > AC

Proof (Step-wise):

1. ∠DBC is an exterior angle at B of triangle ABC.

By the exterior-angle relation,

∠DBC = ∠BAC + ∠ACB

2. Similarly, ∠BCE is an exterior angle at C.

So, ∠BCE = ∠BAC + ∠ABC.

3. Given ∠DBC > ∠BCE,

Substitute the expressions from (1) and (2):

∠BAC + ∠ACB > ∠BAC + ∠ABC

4. Cancel ∠BAC from both sides to get ∠ACB > ∠ABC.

5. In triangle ABC, the larger angle is opposite the larger side.

Since ∠C i.e., ∠ACB > ∠B i.e., ∠ABC, the side opposite ∠C which is AB is greater than the side opposite ∠B which is AC.

Therefore, AB > AC.