Question

In the adjoining figure, ABC is a triangle in which AD is the bisector of ∠A. If AD ⊥ BC, show that ∆ABC is isosceles.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ AD\mathrm{is}\mathrm{a}\mathrm{bisector}\mathrm{of}\angle A. \\ =>\angle DAB=\angle DAC...\left(1\right) \\ AD\perp BC \\ =>\angle BDA=\angle CDA(90°\mathrm{each}) \\ \mathrm{To}\mathrm{prove}: \\ △ABC\mathrm{is}\mathrm{isosceles}. \\ \mathrm{Proof}: \\ \mathrm{In}△DAB\mathrm{and}△DAC: \\ \angle BDA=\angle CDA(90°\mathrm{each}) \\ DA=DA\left(\mathrm{common}\right) \\ \angle \mathrm{DAB}=\angle \mathrm{DAC}(\mathrm{from}1) \\ By\mathrm{ASA}\mathrm{congruence}\mathrm{property}: \\ △DAB\cong △DAC \\ =>AB=AC(\mathrm{corresponding}\mathrm{parts}\mathrm{of}\mathrm{the}\mathrm{congruent}\mathrm{triangles}) \\ \mathrm{Therefore},△\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{is}\mathrm{isosceles}. \end{gathered}$$