Question

In the given figure, $DB\perp BC,DE\perp ABandAC\perp BC.$

Prove that $\frac{BE}{DE}=\frac{AC}{BC}$.

Answer 1

In △BED and △ACB, we have:

∠BED = ∠ACB = 90°
∵ ∠B + ∠C = 180°
∴ BD ∥ AC
∠EBD = ∠CAB (Alternate angles)
Therefore, by AA similarity theorem, we get:
△BED~△ACB
⇒ BE/AC = DE/BC
⇒ BE/DE = AC/BC
This completes the proof.