Question

In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AB² = BC × BD

Answer 1

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