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: $\angle$BAC > $\frac{2}{3}$ of a right angle, i.e., $\angle$BAC > 60$°$

Construct: Mark a point D on side AC such that AD = AB = BD.

Proof:

In $∆$ABD,

$\because$ AD = AB = BD (By construction)

$\therefore \angle 1=\angle 3=\angle 4=60°$

$$\begin{gathered} \mathrm{Now}, \\ \angle BAC=\angle 1+\angle 2=60°+\angle 2 \\ \mathrm{but}60°=\frac{2}{3}\mathrm{of}\mathrm{a}\mathrm{right}\mathrm{angle} \\ \mathrm{So},\angle BAC=\frac{2}{3}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}+\angle 2 \end{gathered}$$

Hence, $\angle$BAC > $\frac{2}{3}$ of a right angle.