Question

In the given figure, ABC is a triangle in which AB = AC. D is a point in the interior of ∆ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC.
Figure

Answer 1

$$\begin{gathered} \mathrm{In}∆BCD, \\ \angle DBC=\angle DCB \\ \mathrm{Therefore},∆DBC\mathrm{is}\text{an}\mathrm{isosceles}\mathrm{triangle}\mathrm{and}\mathrm{the}\mathrm{opposite}\mathrm{sides}\mathrm{of}\mathrm{equal}\mathrm{angles}\mathrm{are}\mathrm{equal}. \\ \therefore DB=DC----\left(1\right) \end{gathered}$$
$$\begin{gathered} AB=AC \\ \text{So},∆ABC\mathrm{is}\mathrm{isosceles}\mathrm{triangle}\mathrm{and}\mathrm{the}\mathrm{opposite}\mathrm{angles}\text{of}\mathrm{the}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}. \\ \therefore \angle ABC=\angle ACB \\ \mathrm{and} \\ \angle DBC=\angle DCB \\ \therefore \angle ABC-\angle DBC=\angle ACB-\angle DCB \\ \therefore \angle ABD=\angle ACD---------\left(2\right) \end{gathered}$$
$$\begin{gathered} \mathrm{And} \\ AB=AC------\left(3\right) \\ \left(\mathrm{given}\right) \end{gathered}$$
From (1), (2) and (3), we have:

$$\begin{gathered} ∆ADB\cong ∆ADC[\mathrm{By}\mathrm{SAS}\mathrm{congruency}\mathrm{criteria}] \\ \therefore \angle BAD=\angle CAD \\ \mathrm{Therefore},AD\mathrm{bisects}\angle BAC. \end{gathered}$$