If ABC is an equilateral  triangle with AD perpendicular to BC, then prove that AD2= 3CD2.

In right ∆ACD,
⇒ AC2 = AD2 + CD2
⇒ BC2 = AD2 + CD2 ( AB = BC = CA)
⇒ (2CD)2 = AD2 + CD2 ( BC = 2CD)
⇒ 4CD2 = AD2 + CD2
⇒ AD2 = 4CD2 – CD2
⇒ AD2 = 3CD2


 

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