Question 3 (i)
In the given figure, ABD is a triangle right angled at A and AC⊥BD. Show that (i)AB2=BC×BD
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Solution
(i) In ΔADB and ΔCAB, we have ∠DAB=∠ACB (Each angle is equal to 90∘) ∠ABD=∠CBA (Common angle) ∴ΔADB∼ΔCAB [AAA similarity criterion] ⇒ABCB=BDAB ⇒AB2=CB×BD
Therefore,AB2=BC×BD