Question Number 10667 by Gaurav3651 last updated on 22/Feb/17 | ||
$${prove}\:{that}\:{the}\:{quadrilateral}\:{formed} \\ $$$${by}\:{angle}\:{bisectors}\:{of}\:{a}\:{cyclic}\: \\ $$$${quadrilateral}\:{is}\:{also}\:{cyclic}. \\ $$ | ||
Answered by mrW1 last updated on 22/Feb/17 | ||
$${ABCD}\:{is}\:{a}\:{convex}\:\:{quadrilateral}. \\ $$$${EFGH}\:{is}\:{the}\:{quadrilateral}\:{which}\:{is} \\ $$$${formed}\:{by}\:{the}\:{angle}\:{bisectors}\:{of} \\ $$$${ABCD}. \\ $$$$ \\ $$$$\angle{A}+\angle{B}+\angle{C}+\angle{D}=\mathrm{360}° \\ $$$$ \\ $$$$\angle{EFG}=\mathrm{180}°−\angle{FAB}−\angle{FBA}=\mathrm{180}°−\frac{\mathrm{1}}{\mathrm{2}}\angle{A}−\frac{\mathrm{1}}{\mathrm{2}}\angle{B} \\ $$$$\angle{EHB}=\mathrm{180}°−\angle{HCD}−\angle{HDC}=\mathrm{180}°−\frac{\mathrm{1}}{\mathrm{2}}\angle{C}−\frac{\mathrm{1}}{\mathrm{2}}\angle{D} \\ $$$$\angle{EFG}+\angle{EHB}=\mathrm{180}°−\frac{\mathrm{1}}{\mathrm{2}}\angle{A}−\frac{\mathrm{1}}{\mathrm{2}}\angle{B}+\mathrm{180}°−\frac{\mathrm{1}}{\mathrm{2}}\angle{C}−\frac{\mathrm{1}}{\mathrm{2}}\angle{D} \\ $$$$=\mathrm{360}°−\frac{\mathrm{1}}{\mathrm{2}}\left(\angle{A}+\angle{B}+\angle{C}+\angle{D}\right)=\mathrm{360}°−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{360}°=\mathrm{180}° \\ $$$$ \\ $$$${in}\:{the}\:{same}\:{way} \\ $$$$\angle{FEH}+\angle{FGH}=\mathrm{180}° \\ $$$$ \\ $$$$\because\:{EFGH}\:{is}\:{cyclic}\:{quadrilateral}. \\ $$$$ \\ $$$${BTW}:\:{RFGH}\:{is}\:{always}\:{cyclic},\:{even}\:{if} \\ $$$${ABCD}\:{is}\:{not}\:{cyclic}! \\ $$ | ||