Question

In ∆ABC, if AD is the bisector of ∠A, show that AB > BD and AC > DC.

Answer 1

$$\begin{gathered} \angle \mathrm{ADB}=\angle \mathrm{CAD}+\angle \mathrm{ACD}(\mathrm{Exterior}\mathrm{angle}\mathrm{property}) \\ \Rightarrow \angle \mathrm{ADB}>\angle \mathrm{CAD} \\ \Rightarrow \angle \mathrm{ADB}>\angle \mathrm{BAD}[\because \angle \mathrm{CAD}=\angle \mathrm{BAD}] \end{gathered}$$
Thus, AB > BD

$$\begin{gathered} \mathrm{Similarly},\angle ADC>\angle BAD \\ \Rightarrow \angle ADC>\angle CAD \end{gathered}$$
Thus, AC > DC
Hence, proved.