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Asked by Aakanksha Last Modified
Answer 
Tasneem
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = 
=
Applying Pythagoras theorem in ΔABE, we obtain
AB2 = AE2 + BE2
4AE2 = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
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Let the side of an equilateral triangle be = a
As altitude bisects the base ,length of altitude will be h =sqrt {a2-(a/2)2}
altitude h =Sqrt (3a2/4)
Now Squaring above both sides : 4h2=3a2
Hence Proved.
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