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