In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC . BD
(ii) AC² = BC . DC
(iii) AD² = BD . CD

(i) Here, ABD is a triangle right angled at A and AC ⊥ BD

⇒ ΔBAD ∼ ΔBCA

⇒ AB/BC = BD/AB (Corresponding sides of similar triangle)

⇒ AB² = BC.BD

(ii) ABD is a triangle right angled at A and AC ⊥ BD

⇒ ΔBCA ∼ ΔACD

⇒ AC/CD= BC/AC

⇒ AC² = BC.DC (Corresponding sides of a similar triangle)

(iii) ABD is a triangle right angled at A and AC ⊥ BD

⇒ ΔBAD ∼ ΔACD

⇒ AD/CD = BD/AD

AD² = BD.CD (Corresponding sides of a similar triangle)