2007-II-49 Posted on 16-06-202121-06-2023 By app.cch No Comments on 2007-II-49 Ans: A Join BC. Consider ΔABD and ΔACB, ∠BAD=∠CAB(common angle)∠ACB=90∘(∠ in semi-circle)∠ABD=90∘(radius ⊥ tangent)∴∠ABD=∠ACB ∠ADB=180∘–∠BAD–∠ABD(∠s sum of Δ)=180∘–∠CAB–∠ACB(proved)=∠ABC(∠s sum of Δ) ∴ΔABD∼ΔACB (A.A.A.). Therefore, we have ADAB=ABAC6AB=AB4AB2=24AB=24=26 Same Topic: 2007-I-17 2008-I-17 2008-II-50 2008-II-51 2007, HKCEE, Paper 2 Tags:Properties of Circles