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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