Question

In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC.
Prove that $\frac{BE}{DE}=\frac{AC}{BC}.$

Answer 1

In $△BEDand△ACB$, we have:

$$\begin{gathered} \angle BED=\angle ACB=90° \\ \because \angle B+\angle C=180° \\ \therefore BD\parallel AC \\ \angle EBD=\angle CAB(\mathrm{Alternate}\mathrm{angles}) \\ \mathrm{Therefore},\mathrm{by}\mathrm{AA}\mathrm{similarity}\mathrm{theorem},\mathrm{we}\mathrm{get}: \\ △BED~△ACB \\ \Rightarrow \frac{BE}{AC}=\frac{DE}{BC} \\ \Rightarrow \frac{BE}{DE}=\frac{AC}{BC} \end{gathered}$$
This completes the proof.