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Question
In triangle ABC, AB = AC and BD is perpendicular to AC.
Prove that: BD2 - CD2 = 2CD × AD
Solution
In right-angled ΔADB,
BD2 = AB2 + AD2 ...(Pythagoras theorem)
⇒ BD2 = AC2 - AD2
⇒ BD2 = (AD + DC)2 - AD2
⇒ BD2 = AD2 + CD2 + 2AD × DC - AD2
⇒ BD2 - DC2 = 2AD × DC
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