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Question
DE || BC and CD || EE Prove that AD2 = AB × AF
Solution
Given: In ∆ABC, DE || BC and CD || EF
To Prove: AD2 = AB × AF
Proof: In ∆ABC, DE || BC ...(Given)
By basic proportionality theorem
`"AB"/"AD" = "AC"/"AE"` ...(1)
In ∆ADC, FE || DC ...(Given)
By basic Proportionality theorem
`"AD"/"AF"= "AC"/"AE"` ...(2)
From (1) and (2) we get
`"AB"/"AD" = "AD"/"AF"`
AD2 = AB × AF
Hence it is proved
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