Question

In the given figure, ∠ACB  90° CD ⊥ AB Prove that `(BC^2)/(AC^2)=(BD)/(AD)`  

 

Answer 1

Given: ∠𝐴𝐶𝐵 = 90° 𝑎𝑛𝑑 𝐶𝐷 ⊥ 𝐴𝐵 

To Prove:` (BC^2)/(AC^2)=(BD)/(AD)` 

Proof: In Δ ACB and Δ CDB  

∠𝐴𝐶𝐵 = ∠𝐶𝐷𝐵 = 90° (𝐺𝑖𝑣𝑒𝑛)
∠𝐴𝐵𝐶 = ∠𝐶𝐵𝐷 (𝐶𝑜𝑚𝑚𝑜𝑛)
By AA similarity-criterion Δ ACB ~ ΔCDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.  

`∴ (BC)/(BD)=(AB)/(BC)`  

`⇒ BC^2=BD.AB` ..............(1) 

In Δ ACB and Δ ADC
∠𝐴𝐶𝐵 = ∠𝐴𝐷𝐶 = 90° (𝐺𝑖𝑣𝑒𝑛) 
∠𝐶𝐴𝐵 = ∠𝐷𝐴𝐶 (𝐶𝑜𝑚𝑚𝑜𝑛)
By AA similarity-criterion Δ ACB ~ ΔADC 

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

∴ `(AC)/(AD)=(AB)/(AC)` 

⇒` AC^2=AD.AB`         .................(2) 

Dividing (2) by (1), we get 

`(BC^2)/(AC^2)=(BD)/(AD)`