Question

Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.(shown in the given figure)

Answer 1

ABCD is a parallelogram.
$$\begin{gathered} \Rightarrow \angle \mathrm{BAD}+\angle \mathrm{CDA}=180°\left(\mathrm{interior}\mathrm{angles}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{transversal}\mathrm{are}\mathrm{supplementary}\right) \\ \Rightarrow \frac{1}{2}\angle \mathrm{BAD}+\frac{1}{2}\angle \mathrm{CDA}=\frac{1}{2}\times 180° \\ \Rightarrow \angle \mathrm{SAD}+\angle \mathrm{SDA}=90°.....\left(1\right) \end{gathered}$$
$$\begin{gathered} \angle \mathrm{SAD}+\angle \mathrm{SDA}+\angle \mathrm{ASD}=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\mathrm{in}△\mathrm{ASD}\right) \\ \Rightarrow 90°+\angle \mathrm{ASD}=180° \\ \Rightarrow \angle \mathrm{ASD}=90° \\ \Rightarrow \angle \mathrm{PSR}=90°\left(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right) \end{gathered}$$
Similarly, $\angle \mathrm{PQR}=90°$
$$\begin{gathered} \Rightarrow \angle \mathrm{DAB}+\angle \mathrm{CBA}=180°\left(\mathrm{interior}\mathrm{angles}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{transversal}\mathrm{are}\mathrm{supplementary}\right) \\ \Rightarrow \frac{1}{2}\angle \mathrm{DAB}+\frac{1}{2}\angle \mathrm{CBA}=\frac{1}{2}\times 180° \\ \Rightarrow \angle \mathrm{BAR}+\angle \mathrm{RBA}=90° \\ \mathrm{In}△\mathrm{ABR}, \\ \angle \mathrm{BAR}+\angle \mathrm{RBA}+\angle \mathrm{ARB}=180° \\ \Rightarrow 90°+\angle \mathrm{ARB}=180° \\ \Rightarrow \angle \mathrm{ARB}=90° \\ Similarly, \\ \angle \mathrm{DPC}=90° \end{gathered}$$
Thus, PQRS is a rectangle.