Question

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

Answer 1

In ∆BED and ∆ACB, we have: 

∠BED = ∠ACB = 90° 

∵ ∠B + ∠C = 180°

∴ BD || AC 

∠EBD = ∠CAB (alternative angles ) 

Therefore, by AA similarity theorem, we get : 

∆ BED ~ ∆ ACB 

⇒ BE/AC = DE/BC 

⇒ BE/DE = AC/BC

This completes the proof.