Question

In the given figure, BA ⊥ AC and DE ⊥ EF such that BA =DE and BF = DC. Prove that AC = EF.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}BF=DC \\ \Rightarrow BF+CF=DC+CF \\ \Rightarrow BC=DF...\left(i\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In}△\mathrm{ABC}\mathrm{and}△\mathrm{EDF},\mathrm{we}\mathrm{have}: \\ \mathrm{BA}=\mathrm{DE}\left(\mathrm{Given}\right) \\ \angle \mathrm{BAC}=\angle \mathrm{DEF}(90°\mathrm{each}) \\ \mathrm{BC}=\mathrm{DF}\left[\mathrm{From}\right(\mathrm{i}\left)\right] \\ \mathrm{Thus},△\mathrm{ABC}\cong △\mathrm{EDF}(\mathrm{RHS}\mathrm{criterion}) \\ \therefore \mathrm{AC}=\mathrm{EF}\left(\mathrm{CPCT}\right) \end{gathered}$$
Hence, proved.