**ABC is an equilateral triangle of side 2a. **

**Draw, AD ⊥ BC**

**In ΔADB and ΔADC, we have**

**AB = AC [Given]**

**AD = AD [Given]**

**∠ADB = ∠ADC [equal to 90°]**

**Therefore, ΔADB ≅ ΔADC by RHS congruence.**

**Hence, BD = DC [by CPCT]**

**In right angled ΔADB, **

**AB**^{2} = AD^{2} + BD^{2}

**(2***a*)2 = AD2 + *a*2

**⇒ AD**^{2} = 4*a*^{2} - *a*^{2}

**⇒ AD**^{2} = 3*a*^{2}

**⇒ AD = √3a **