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.
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Answer
Given: In $\triangle\text{ABC, BC}$ is the longest side. To prove: $\angle\text{BAC}>\frac{2}{3}$ of a right angle, i.e., $\angle\text{BAC}>60^{\circ}$ Construct: Mark a point D on side AC such that AD = AB = BD. Proof: In $\triangle\text{ABD,}$$\because\text{AD = AB = BD}$ (By construction)$\therefore\angle1=\angle3=\angle4=60^{\circ}$
Now,$\angle\text{BAC}=\angle1+\angle2=60^{\circ}+\angle2$
but $60^{\circ}=\frac{2}{3}$ of a right angle So, $\angle\text{BAC}=\frac{2}{3}$ of a right angle + $\angle2$ Hence, $\angle\text{BAC}>\frac{2}{3}$ of a right angle.
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