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Question
In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AD2 = BD × CD
Solution
In ΔDCA and ΔDAB, we have
∠DCA = ∠DAB (Each equals to 90°)
∠CDA = ∠ADB (common angle)
∴ ΔDCA ~ ΔDAB [By AA similarity criterion]
`⇒ (DC)/(DA) = (DA)/(DA)`
⇒ AD2 = BD × CD
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