In the given figure, ∠ACB = 90° CD ⊥ AB Prove that BC2/AC2 =

1.	In the given figure, ∠ACB = 90° CD ⊥ AB Prove that BC2/AC2 = BD/AD
Answer» Given: ∠ACB = 90° And CD⊥AB To Prove; BC²/AC²= BD/AD Proof: In ∆ ACB and ∆ CDB ∠ACB = ∠COB = 90° (Given) ∠ = ∠ (Common) By AA similarity-criterion ∆ ACB ~ ∆CDB When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional. ∴ BC/BD = AB/BC ⇒ BC² = BD. AB … (1) In ∆ ACB and ∆ ADC ∠ACB = ∠ADC = 90° (Given) ∠CAB = ∠DAC (Common) By AA similarity-criterion ∆ ACB ~ ∆ADC When two triangles are similar, then the ratios of their corresponding sides are proportional. ∴ AC/AD = AB/AC ⇒ AC² = AD. AB ….(2) Dividing (2) by (1), we get BC²/AC² = BD/AD

In the given figure, ∠ACB = 90° CD ⊥ AB Prove that BC2/AC2 = BD/AD

Answer»

Given: ∠ACB = 90° And CD⊥AB

To Prove; BC²/AC²= BD/AD

Proof: In ∆ ACB and ∆ CDB

∠ACB = ∠COB = 90° (Given)

∠ = ∠ (Common)

By AA similarity-criterion ∆ ACB ~ ∆CDB

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.

∴ BC/BD = AB/BC

⇒ BC² = BD. AB … (1)

In ∆ ACB and ∆ ADC

∠ACB = ∠ADC = 90° (Given)

∠CAB = ∠DAC (Common)

By AA similarity-criterion ∆ ACB ~ ∆ADC

When two triangles are similar, then the ratios of their corresponding sides are proportional.

∴ AC/AD = AB/AC

⇒ AC² = AD. AB ….(2)

Dividing (2) by (1), we get

BC²/AC² = BD/AD