Q14 of 95 Page 5

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


and AD ⊥ BC


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


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


⇒ (AD)2 + (BD)2 = (AB)2


⇒ (AD)2 + (BD)2 = (BC)2 [∵ AB = BC]


⇒ (AD)2 + (BD)2 = (2BD)2 [ as AD⊥BC]


⇒ (AD)2 + (BD)2 = 4BD2


⇒ AD2 = 3BD2


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