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In an isosceles triangle ABC with AB = AC, BD is perpendicular from B to side AC. Prove that BD2 – CD2 = 2 CD.AD

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Best answer

Since ΔADB is right angled at D,

AB2 = AD2 + BD2

⇒ AC2 = AD2 + BD2       (∴AB = AC) 

⇒ (AD+CD)2 = AD2 + BD2 

⇒ AD2 + CD2 + 2AD.CD = AD2 + BD2 

⇒ BD2 – CD2 = 2AD.CD

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