Question

In ∆ABC angle BAC= 90°, AD is perpendicular to BC. Prove that AD/BD= AD/CD= BD×CD​

Answer 1

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!