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Question
In ∆ABC, given that AB = AC and BD ⊥ AC. Prove that BC2 = 2 AC. CD
Solution
Since Δ ADB is right triangle right angled at D
`AB^2=AD^2+BD^2`
Substitute `AB=AC`
`AC^2 =AD^2+BD^2`
`AC^2=(AC-DC)^2+BD^2`
`AC^2=AC^2+DC^2-2AC.DC+BD^2`
`2AC.DC=AC^2-AC^2+DC^2+BD^2`
`2AC.DC=DC^2+BD^2`
Now, in Δ BDC, we have
`CD^2+BD^2=BC^2`
Therefore , `2AC.DC = DC^2+BD^2`
`2AC.DC=BC^2`
Hence proved.
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