Given: ABC is an equilateral triangle
∴ AB = AC = BC = 2a
And let AD is an altitude on BC. Therefore, BD = 1/2 x BC = a
Now, In ∆ADB, using Pythagoras theorem, we have
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AD)2 + (BD)2 = (AB)2
⇒ (AD)2 + (a)2 = (2a)2
⇒ (AD)2 = 4a2 – a2
⇒ (AD)2 = 3a2
⇒ AD = a√3 units