prove that in a triangle other the than equilateral triangle opposite to longer side is greater than 2/3 of right angle
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Given : ΔABC other than equilateral triangle
Let AB = AD = DB
ABD is an equilateral triangle since AB = AD = DB
∠ABD = ∠ADB = ∠BAD = 60°
Longest side= BC
Angle opposite to longest side = ∠BAC
Therefore it is clear that ∠BAC = ∠BAD + ∠CAD
So ∠BAC > ∠BAD
∠BAC > 60°
Hence proved
Let AB = AD = DB
ABD is an equilateral triangle since AB = AD = DB
∠ABD = ∠ADB = ∠BAD = 60°
Longest side= BC
Angle opposite to longest side = ∠BAC
Therefore it is clear that ∠BAC = ∠BAD + ∠CAD
So ∠BAC > ∠BAD
∠BAC > 60°
Hence proved
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