Answer :

We have ABC is an equilateral triangle and ADBC


In ADB ADC


ADB=ADC=90°
AB=AC (Given)


AD=AD (Common)


ADB≅⊿ ADC (By RHS condition)


BD=CD=BC/2 ……. (i)


In ABD


BC2=AD2+BD2


BC2=AD2+BD2 [Given AB=BC]


(2BD)2= AD2+BD2 [From (i)]


4BD2-BD2=AD2


AD2=3BD2


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