Question

In the following figure, angleB=90°, BD is perpendicular to AC. prove that BC^2/AB^2=DC/AD​

Answer 1

DC/AD = BC²/AB²

Step-by-step explanation:

in Δ ADB  & Δ ABC

∠A = ∠A  ( common)

∠ADB = ∠ABC = 90°

=> Δ ADB  ≈ Δ ABC

=> AD/AB  = BD/BC  = AB/AC

AD/AB   = AB/AC

=> AD * AC = AB²

=> AC = AB²/AD

Similarly in Δ CDB  & Δ CBA

∠C = ∠C  ( common)

∠CDB = ∠CBA = 90°

=> Δ CDB  ≈ Δ CBA

=> DC/BC  = BD/BA  = BC/AC

DC/BC  =   = BC/AC

=> DC* AC = BC²

=> AC = BC²/DC

Equating AC

AB²/AD = BC²/DC

=> DC/AD = BC²/AB²

QED

proved

Learn more:

in a trianle abc angle b is 90 degree bd perpendicular to ac. bc is 8 ...

https://brainly.in/question/7297531

In triangle abc angle b is equal to 90 degree and bd is perpendicular ...

https://brainly.in/question/12244192