Question

Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle.

Answer 1

Given: In ∆ ABC, BC is the longest side.
To prove: ∠ BAC > 2/3 of a right angle, i.e., ∠ BAC > 60°
Construct: Mark a point D on side AC such that AD = AB = BD.
Proof:
In ∆ ABD,
∵ AD = AB = BD (By construction)
∴∠1=∠3=∠4=60°
Now,∠BAC=∠1+∠2=60°+∠2but 60° =2/3 of a right angleSo, ∠BAC=2/3 of a right angle + ∠2
Hence, ∠ BAC > 2/3 of a right angle.