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Question

In triangle ABC, AC=AB and BD is perpendicular to AC prove that BD2+CD2=2AC×CD

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Solution


Given: BD is perpendicular to AC and AB = AC
To prove: BC2=2AC.CD
Proof: In ΔBCD by pythagoras theorem, we have
BC2=BD2+CD2(1)
Again, in ΔABD by pythagoras theorem
AB2=BD2+AD2
BD2=AB2AD2
On putting value of BD2 in (1), we get
BC2=AB2AD2+CD2
BC2=AB2(ACCD)2+CD2
BC2=AB2AC2CD2+2AC.CD+CD2
BC2=2AC.CD [Given, AB=ACAB2=AC2]

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