Question

In ∆ABC, BD ⊥ AC and CE ⊥ AB such that BE = CD. Prove that BD = CE.

Answer 1

In $△BDCand△CEB,$ we have:
$$\begin{gathered} \mathrm{BE}=\mathrm{CD}\left(\mathrm{Given}\right) \\ \angle \mathrm{BEC}=\angle \mathrm{CDB}=90° \\ \Rightarrow \mathrm{BC}=\mathrm{BC}\left(\mathrm{Common}\right) \\ △\mathrm{BDC}\cong △\mathrm{CEB}(\mathrm{RHS}\mathrm{criterion}) \end{gathered}$$
∴ $BD=CE$ (CPCT)
Hence proved.