Explanation
{\textbf{Step-1: Drawing the diagram as per given question and proving the required statement}}
{\text{Here, ABCD is a cyclic quadrilateral}}
{\text{AH, BF, CF and DH are the angle bisectors of}} \angle {\text{A,}} \angle {\text{B,}} \angle {\text{C,}} \angle {\text{D.}}
\angle {\text{FEH =}} \angle {\text{AEB ……. (1) [Vertically opposite angles]}}
\angle {\text{FGH =}} \angle {\text{DGC.…… (2) [Vertically opposite angles]}}
{\text{Adding (1) and (2),}}
\angle {\text{FEH +}} \angle {\text{FGH =}} \angle {\text{AEB + DGC …..(3)}}
{\text{Now, by angle sum property of a triangle,}}
\angle {\text{AEB =}} {\text{18}}{{\text{0}}^{\text{0}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{A + }}\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{B}}} \right) {\text{…….(4)}}
\angle {\text{DGC = 18}}{{\text{0}}^{\text{0}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{C + }}\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{D}}} \right) {\text{………..(5)}}
{\text{Substituting (4) and (5) in equation (3)}}
\angle {\text{FEH + }}\angle {\text{FGH = 18}}{{\text{0}}^{\text{0}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{A + }}\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{B}}} \right){\text{ + 18}}{{\text{0}}^{\text{0}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{C + }}\dfrac{{\text{1}}}{{\text{2}}}\angle {\text{D}}} \right)
\angle {\text{FEH + }}\angle {\text{FGH = 36}}{{\text{0}}^{\text{0}}}{\text{ - }}\dfrac{1}{2}\left( {\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C + }}\angle {\text{D}}} \right)
\angle {\text{FEH + }}\angle {\text{FGH = 36}}{{\text{0}}^{\text{0}}}{\text{ - }}\dfrac{1}{2} \times {360^0}
\angle {\text{FEH + }}\angle {\text{FGH = 18}}{{\text{0}}^0}
{\text{Now, the sum of opposite angles of quadrilateral EFGH is}} {\text{18}}{{\text{0}}^0}
{\text{EFGH is a cyclic quadrilateral}}
{\textbf{ Hence, the quadrilateral formed by angle bisectors of a cyclic quadrilateral}}
{\textbf{is also cyclic.}}
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