Question

 In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that AD² = BD × CD

Answer 1

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