Question

In equilateral ∆ABC, AD ⟂ BC. Prove that 3 BC² = 4AD²

Answer 1

PYTHAGORAS THEOREM: In a right angle triangle the square of the hypotenuse is equal to the sum of the square of the other two sides.

GIVEN:
A equilateral ∆ ABC, in which sides are AB=BC= AC= a units and AD ⟂ BC ,
In ∆ADB ,
AB²= AD²+ BD²     (by Pythagoras theorem)
a² = AD² + (a/2)²   

[BD= 1/2BC, since in an equilateral triangle altitude AD is  ⟂ bisector of BC ]
a²- a²/4 =AD²
( 4a²-a²)/4 = AD²
3a² /4 = AD²
3BC²/4= AD²
[ BC= a]
3BC²= 4AD²

Hence, proved

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Answer 2

Hey...!!! :))
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given-  ABC is an equilateral triangle

to prove that - 3BC2=4AD2 

proof - by pythagoras theorem in triangle ABD

          BC2  = AD2 + BD2

            but BD = 1/2 AB

thus BC2 = AD2 + {1/2 BC}2

  BC2 = AD2 + 1/4 AB2

4BC2 = 4AD2 + AB2

4  BC2 - AB2 = 4 AD2

thus 3BC2=4AD2 { as AB =BC we can subtract them}
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