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Question
In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AB2 = BC × BD
Solution
In ΔADB and ΔCAB, we have
∠DAB = ∠ACB (Each equals to 90°)
∠ABD = ∠CBA (Common angle)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
`⇒(AB)/(CB) = (BD)/(AB)`
⇒ AB2 = CB × BD
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