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Question

In an equilateral triangle ABC, AD is parpendicular to BC . Prove that 3AB2 = 4AD2.


Solution

in triangle ABC,

sides are 'a' units

as AD is perpendicular to BC ,

then in triangle ABD ,

AB2 = AD2 + BD2

a2 = AD2 + (a/2)2   [ since AD is perpendicular to BC ]

a2 - a2/4 =AD2

= 3a2 /4 = AD2 ..........(1)

in triangle ADC, 

AC2 = AD2 + CD2

a2 = AD2 + (a/4)2

3a2/4 = AD2 .........(2)

from equation 1 and 2

3 a2 /4 = AD2

3 a2 = 4 AD2

3 AB2 = 4AD2 [since AB = a ]

hence proved 

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