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Question
In ∆ABC, if BD ⊥ AC and BC2 = 2 AC . CD, then prove that AB = AC.
Solution
Since Δ ADB is right triangle right angled at D
`AB^2=AD^2+BD^2`
In right Δ BDC, we have
`CD^2+BD^2=BC^2`
Sine `2AC.DC=BC^2`
`⇒DC^2+BD^2=2AC.DC `
`2AC.DC=AC^2-AC^2+DC^2+BD^2`
`AC^2=AC^2+DC^2-2AC.DC+BD^2`
`AC^2=(AC-DC)^2+BD^2`
`AC^2=AD^2+BD^2`
Now substitute `AD^2+BD^2=AB^2`
`AC^2=AB^2`
`AC=AB`
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