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Question
In triangle ABC, AB = AC and BD is perpendicular to AC.
Prove that: BD2 − CD2 = 2CD × AD
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Solution
Use the Pythagoras Theorem in triangle BDC
BD2 = BC2 − CD2
BC = CD + AD ⇒ BC2 = (CD + AD)2 = CD2 + 2CD ⋅ AD + AD2
BD2 = BC2 − CD2 = (CD2 + 2CD ⋅ AD + AD2) − CD2 = 2CD ⋅ AD + AD2
Now subtract CD2:
BD2 − CD2 = 2CD ⋅ AD
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