prove that in a triangle other than an equilateral triangle, angle opposite the longest side is grater than 2/3 of a right angle,,
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It follows from the law of sines that the angle opposite to the longest side will be the largest angle. By the Pigeonhole principle, if we were to distribute 180 "degrees" randomly into three "angles," at least one of the angles would have to be at least 60 degrees. However, the largest angle cannot be exactly 60 degrees, so it must be more than 60 degrees.
The construction is already given in the picture.So let me just establish the relation between the sides.
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
Here it is clear that ∠BAC = ∠BAD + ∠CAD
So ∠BAC > ∠BAD
∠BAC > 60°
Hence proved!
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