Question

If all the sides of a quadrilateral are produced in the same order (clockwise or anticlockwise).
Show that quadrilateral formed by the bisectors of exterior angles is cyclic.

Answer 1

$$\begin{gathered} \angle \mathrm{EAD}+\angle 1=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow 2\angle 6+\angle 1=180°(\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle \mathrm{EAD}) \\ \Rightarrow \angle 6=90°-\frac{\angle 1}{2} \\ \angle \mathrm{FBC}+\angle 2=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow 2\angle 7+\angle 2=180°(\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle \mathrm{FBC}) \\ \Rightarrow \angle 7=90°-\frac{\angle 2}{2} \\ \angle \mathrm{GCB}+\angle 3=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow 2\angle 8+\angle 3=180°(\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle \mathrm{GCB}) \\ \Rightarrow \angle 8=90°-\frac{\angle 3}{2} \\ \angle \mathrm{SDA}+\angle 4=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow 2\angle 5+\angle 4=180°(\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle \mathrm{SDA}) \\ \Rightarrow \angle 5=90°-\frac{\angle 4}{2} \end{gathered}$$


$$\begin{gathered} \mathrm{Now},\angle 5+\angle 6+\angle 10=180°...\left(\mathrm{i}\right)(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \angle 7+\angle 8+\angle 9=180°...\left(\mathrm{ii}\right)(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \mathrm{Adding}\mathrm{equations}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{have}: \\ \angle 5+\angle 6+\angle 10+\angle 7+\angle 8+\angle 9=360° \\ \Rightarrow \angle 9+\angle 10+\angle 5+\angle 6+\angle 7+\angle 8=360° \\ \Rightarrow \angle 9+\angle 10+\left(90°-\frac{\angle 4}{2}\right)+\left(90°-\frac{\angle 1}{2}\right)+\left(90°-\frac{\angle 2}{2}\right)+\left(90°-\frac{\angle 3}{2}\right)=360° \\ \Rightarrow \angle 9+\angle 10+360°-\frac{1}{2}\left(\angle 1+\angle 4+\angle 2+\angle 3\right)=360° \\ \Rightarrow \angle 9+\angle 10+360°-\frac{1}{2}\times 360°=360° \\ \Rightarrow \angle 9+\angle 10+180°=360° \\ \Rightarrow \angle 9+\angle 10=180° \end{gathered}$$

Hence, quad PQRS is cyclic.