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Question Number 10667 by Gaurav3651 last updated on 22/Feb/17

prove that the quadrilateral formed  by angle bisectors of a cyclic   quadrilateral is also cyclic.

$${prove}\:{that}\:{the}\:{quadrilateral}\:{formed} \\ $$$${by}\:{angle}\:{bisectors}\:{of}\:{a}\:{cyclic}\: \\ $$$${quadrilateral}\:{is}\:{also}\:{cyclic}. \\ $$

Answered by mrW1 last updated on 22/Feb/17

ABCD is a convex  quadrilateral.  EFGH is the quadrilateral which is  formed by the angle bisectors of  ABCD.    ∠A+∠B+∠C+∠D=360°    ∠EFG=180°−∠FAB−∠FBA=180°−(1/2)∠A−(1/2)∠B  ∠EHB=180°−∠HCD−∠HDC=180°−(1/2)∠C−(1/2)∠D  ∠EFG+∠EHB=180°−(1/2)∠A−(1/2)∠B+180°−(1/2)∠C−(1/2)∠D  =360°−(1/2)(∠A+∠B+∠C+∠D)=360°−(1/2)×360°=180°    in the same way  ∠FEH+∠FGH=180°    ∵ EFGH is cyclic quadrilateral.    BTW: RFGH is always cyclic, even if  ABCD is not cyclic!

$${ABCD}\:{is}\:{a}\:{convex}\:\:{quadrilateral}. \\ $$$${EFGH}\:{is}\:{the}\:{quadrilateral}\:{which}\:{is} \\ $$$${formed}\:{by}\:{the}\:{angle}\:{bisectors}\:{of} \\ $$$${ABCD}. \\ $$$$ \\ $$$$\angle{A}+\angle{B}+\angle{C}+\angle{D}=\mathrm{360}° \\ $$$$ \\ $$$$\angle{EFG}=\mathrm{180}°−\angle{FAB}−\angle{FBA}=\mathrm{180}°−\frac{\mathrm{1}}{\mathrm{2}}\angle{A}−\frac{\mathrm{1}}{\mathrm{2}}\angle{B} \\ $$$$\angle{EHB}=\mathrm{180}°−\angle{HCD}−\angle{HDC}=\mathrm{180}°−\frac{\mathrm{1}}{\mathrm{2}}\angle{C}−\frac{\mathrm{1}}{\mathrm{2}}\angle{D} \\ $$$$\angle{EFG}+\angle{EHB}=\mathrm{180}°−\frac{\mathrm{1}}{\mathrm{2}}\angle{A}−\frac{\mathrm{1}}{\mathrm{2}}\angle{B}+\mathrm{180}°−\frac{\mathrm{1}}{\mathrm{2}}\angle{C}−\frac{\mathrm{1}}{\mathrm{2}}\angle{D} \\ $$$$=\mathrm{360}°−\frac{\mathrm{1}}{\mathrm{2}}\left(\angle{A}+\angle{B}+\angle{C}+\angle{D}\right)=\mathrm{360}°−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{360}°=\mathrm{180}° \\ $$$$ \\ $$$${in}\:{the}\:{same}\:{way} \\ $$$$\angle{FEH}+\angle{FGH}=\mathrm{180}° \\ $$$$ \\ $$$$\because\:{EFGH}\:{is}\:{cyclic}\:{quadrilateral}. \\ $$$$ \\ $$$${BTW}:\:{RFGH}\:{is}\:{always}\:{cyclic},\:{even}\:{if} \\ $$$${ABCD}\:{is}\:{not}\:{cyclic}! \\ $$

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