Question 3 (iii)
In the given figure, ABD is a triangle right angled at A and AC⊥BD. Show that (iii)AD2=BD×CD
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Solution
(iii) In ΔDCAandΔDAB, we have ∠DCA=∠DAB (Each angle is equal to 90∘) ∠CDA=∠ADB (common angle) ∴ΔDCA∼ΔDAB [By AAA similarity criterion] ⇒DCDA=DADB ⇒AD2=BD×CD