Question

If the bisector of the vertical angle of a triangle bisects the base, prove that the triangle is isosceles.

Answer 1

$$\begin{gathered} \mathrm{In}∆ABC, \\ \angle BAD=\angle CAD \\ \mathrm{and}BD=CD \\ \mathrm{Extend}AD\mathrm{to}\mathrm{point}E\text{so}\mathrm{that}CE\mathrm{is}\mathrm{parallel}\mathrm{to}AB. \\ \mathrm{Join}CE. \\ \mathrm{Taking}\mathrm{transversal}AE\mathrm{intersecting}AB\mathrm{and}CE, \\ \angle BAD=\angle DEC\left(\mathrm{interior}\mathrm{alternate}\mathrm{angle}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that}\angle BAD=\angle CAD=\angle DEC \\ \mathrm{In}△ACE, \\ \angle CAD\hspace{0.17em}=\angle DEC \\ \therefore AC\hspace{0.17em}=CE \\ \mathrm{In}△ABD\mathrm{and}△ECD, \\ \angle ADB=\angle EDC\left(\mathrm{vertically}-\mathrm{opposite}\mathrm{angle}\right) \\ \angle BAD=\angle CED\left(\mathrm{interior}\mathrm{alternate}\mathrm{angle}\right) \\ BD=CD\left(\mathrm{given}\right) \\ \mathrm{By}AAS,∆ABD\cong △ECD \\ \mathrm{Therefore}, \\ AB=EC\left(\mathrm{by}\mathrm{cpct}\right) \\ \mathrm{We}\mathrm{know}EC=AC\left(\mathrm{already}\mathrm{proven}\right) \\ \mathrm{Therefore}, \\ AB=AC\mathrm{and}△ABC\mathrm{is}\mathrm{isosceles}. \end{gathered}$$