Question

In the given figure, ABC is a triangle in which AB = AC. If D be a point on BC produced, prove that AD > AC.

Answer 1

$$\begin{gathered} AB=AC \\ \angle ABC=\angle ACB \end{gathered}$$

$$\begin{gathered} \mathrm{As}\angle ACB>\angle ADC,\mathrm{we}\mathrm{have}: \\ \angle ABC>\angle ADB \\ \Rightarrow \angle ABD>\angle ADB \end{gathered}$$

∴ AD > AB

AB =AC (Given)
∴ AD > AC
Hence, proved.