Question

In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that (i) AC bisects ∠A and ∠C, (ii) AC is the perpendicular bisector of BD.

Answer 1

Consider the triangles ABC and ADC.
$$\begin{gathered} \mathrm{AB}=\mathrm{AD}\left(\mathrm{Given}\right) \\ \mathrm{BC}=\mathrm{BD}\left(\mathrm{Given}\right) \\ \mathrm{AC}=\mathrm{AC}\left(\mathrm{Common}\right) \\ \therefore △\mathrm{ABC}\cong △\mathrm{ADC}(\mathrm{SSS}\mathrm{criterion}) \\ \Rightarrow \angle \mathrm{BAC}=\angle \mathrm{DAC}(\mathrm{Corresponding}\mathrm{angles}\mathrm{of}\mathrm{congruent}\mathrm{triangles}) \\ \Rightarrow \angle \mathrm{BCA}=\angle \mathrm{DCA}(\mathrm{Corresponding}\mathrm{angles}\mathrm{of}\mathrm{congruent}\mathrm{triangles}) \end{gathered}$$

Thus, AC bisects angles A and C.

Let AC intersect BD at O. Consider the triangles AOB and AOD.
$$\begin{gathered} \angle \mathrm{BAO}=\angle \mathrm{DAO}\left(\mathrm{Proved}\right) \\ \mathrm{AB}=\mathrm{AD}\left(\mathrm{Given}\right) \\ \mathrm{Thus},△\mathrm{ABD}\hspace{0.17em}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{traingle}. \\ \mathrm{i}.\mathrm{e}.,\angle \mathrm{ABO}=\angle \mathrm{ADO} \\ \Rightarrow △\mathrm{AOB}\cong △\mathrm{AOD}(\mathrm{AAS}\mathrm{criterion}) \\ \Rightarrow \mathrm{BO}=\mathrm{OD} \end{gathered}$$
Thus, AC bisects BD at O.

$$\begin{gathered} \mathrm{Further},\angle \mathrm{BOA}=\angle \mathrm{DOA}(\mathrm{Corresponding}\mathrm{angles}\mathrm{of}\mathrm{congruent}\mathrm{triangles}) \\ \mathrm{Also},\angle \mathrm{BOA}+\angle \mathrm{DOA}=180° \\ \Rightarrow \angle \mathrm{BOA}=\angle \mathrm{DOA}=90° \end{gathered}$$

Thus, AC is the perpendicular bisector of BD.