# In an equilateral triangle ABC, AD is drawn perpendicular to BC, meeting BC in D. Prove that AD2 = 3BD2.

Given: ABC is an equilateral triangle

AB = AC = BC

Now, In ∆ADB, using Pythagoras theorem, we have

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

(AD)2 + (BD)2 = (BC)2 [ AB = BC]

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