Question

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

Answer 1

Prove that the triangle is isosceles

Given, in $∆ABC,AD$ is a bisector of $\angle A$ and meet base $BC$at $D$, such that

$BD=DC$

Construction: Produce $AD$ to meet $E$ such that $AD=ED$

Now, in $∆ABC$ and $∆DEC,$

$$\begin{gathered} BD=DC......................\left[\mathrm{Given}\right] \\ AD=DE......................\left[\mathrm{By}\mathrm{construction}\right] \\ \angle ADB=\angle EDC..........\left[\mathrm{Vertically}\mathrm{opposite}\mathrm{angle}\right] \\ \therefore ∆ABD\cong ∆EDC.....\left[BySAS\mathrm{congruence}\right] \end{gathered}$$

Thus by CPCT rule,

$AB=EC$ and $\angle BAD=\angle DEC$

Also, $\angle BAD=\angle DAC$

In $∆ACE,\angle AEC=\angle CAE$

$$\begin{gathered} AC=CE......................\left[\mathrm{Sides}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{angle}\right] \\ \Rightarrow AB=AC \end{gathered}$$

Hence, $∆ABC$ is isosceles