Question

In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that `(BE)/(DE) = (AC)/(BC)`.

 

Answer 1

In ΔBED and ΔACB, we have:
∠𝐵𝐸𝐷= ∠𝐴𝐶𝐵=90°
∵ ∠𝐵+ ∠𝐶=180°
∴ BD || AC
∠𝐸𝐵𝐷= ∠𝐶𝐴𝐵 (𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 )
Therefore, by AA similarity theorem, we get :
Δ BED ~ Δ ACB 

⇒` (BE)/(AC)=(DE)/(BC)` 

⇒ `(BE)/(DE)=(AC)/(BC)` 

This completes the proof.