#### Question

In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AB^{2} = 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)`

⇒ AB^{2} = CB × BD

Is there an error in this question or solution?

Solution In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AB^2 = BC × BD Concept: Pythagoras Theorem.