In an equilateral, , prove that .

We have _{⊿ ABC is an equilateral triangle and AD⊥BC}

_{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