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

(N/A) Given: $A$ triangle $ABC$,which is not an equilateral triangle. Let $BC$ be the longest side.
To prove: $\angle A > \frac{2}{3} \times 90^{\circ} = 60^{\circ}$.
Proof: In $\Delta ABC$,since $BC$ is the longest side,we have:
$BC > AB \Rightarrow \angle A > \angle C$ ..... $(1)$ [Since the angle opposite to the longer side is larger]
$BC > AC \Rightarrow \angle A > \angle B$ ..... $(2)$ [Since the angle opposite to the longer side is larger]
Adding $(1)$ and $(2)$,we get:
$\angle A + \angle A > \angle B + \angle C$
$2\angle A > \angle B + \angle C$
Adding $\angle A$ on both sides:
$2\angle A + \angle A > \angle A + \angle B + \angle C$
$3\angle A > 180^{\circ}$ [Angle sum property of a triangle]
$\angle A > \frac{180^{\circ}}{3}$
$\angle A > 60^{\circ}$
Since $60^{\circ} = \frac{2}{3} \times 90^{\circ}$,we have $\angle A > \frac{2}{3}$ of a right angle.
Hence,proved.