Question

ABC is a triangle in which AB = AC. If the bisectors of ∠B and ∠C meet AC and AB in D and E respectively, prove that BD = CE.

Answer 1

In triangle ABC, we have:
AB = AC (Given)
$$\begin{gathered} \Rightarrow \angle \mathrm{B}=\angle \mathrm{C} \\ \Rightarrow \frac{1}{2}\angle \mathrm{B}=\frac{1}{2}\angle \mathrm{C} \\ \therefore \angle \mathrm{ABD}=\angle \mathrm{ACE}...\left(\mathrm{i}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{in}△\mathrm{ADB}\mathrm{and}△\mathrm{ACE},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{ABD}=\angle \mathrm{ACE}[\mathrm{From}(\mathrm{i}\left)\right] \\ \mathrm{AB}=\mathrm{BC}\left(\mathrm{Given}\right) \\ \angle \mathrm{BAD}=\angle \mathrm{CAE}(\mathrm{Common}\mathrm{angle}) \\ \therefore △\mathrm{ADB}\cong △\mathrm{AEC}(\mathrm{ASA}\mathrm{criterion}) \end{gathered}$$

∴ $BD=CE(\mathrm{Corresponding}\mathrm{sides}\mathrm{of}\mathrm{congruent}\mathrm{triangles})$
Hence, proved.