In ∆ABC angle BAC= 90°, AD is perpendicular to BC. Prove that AD/BD= AD/CD= BD×CD
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Answer:
In △ABC, if ∠ABC=θ, ∠ACB=(90−θ)
∴ in △ACD,∠CAD=90−(90−θ)=θ
Refer triangle, △ABD
tanθ=AD/BD-----(1)
Refer △ACD,
tanθ=CD/AD-----(2)
Equating (1),(2)⟹ Tanθ=AD/BD=CD/AD
AD²=BD×CD
Hence proved!
hope this helps!
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