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Question
In the adjoining figure, ΔADB ∼ ΔBDC. Prove that BD2 = AD × DC.
Solution
Given: ΔADB ∼ ΔBDC
∴ By the corresponding side's criterion of similarity,
`("AD")/("BD") = ("BD")/("DC")`
⇒ AD × DC = BD × BD
⇒ BD2 = AD × DC
Hence proved.
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