Question Number 42352 by Raj Singh last updated on 24/Aug/18
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18
$$\angle{AOC}=\angle{BOD}=\mathrm{2}{x}\:\left({vertically}\:{opposite}\:{angle}\:{voa}\right) \\ $$$$\angle{AOB}=\angle{COD}\:\:=\mathrm{2}{y}\left({voa}\right) \\ $$$$\mathrm{2}{x}+\mathrm{2}{y}+\mathrm{2}{x}+\mathrm{2}{y}=\mathrm{360}^{{o}} \\ $$$${x}+{y}+{x}+{y}=\mathrm{180}^{{o}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\angle{AOC}\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\angle{BOD} \\ $$$$\mathrm{2}{y}=\angle{AOB} \\ $$$${so}\:{x}+\mathrm{2}{y}+{x}=\mathrm{180}^{{o}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\angle{AOC}+\angle{AOB}+\frac{\mathrm{1}}{\mathrm{2}}\angle{BOD}=\mathrm{180}^{{o}} \\ $$$$\angle{POA}+\angle{AOB}+\angle{BOQ}=\mathrm{180}^{{o}} \\ $$$${so}\:{OP}\:{and}\:{OQ}\:{on}\:{the}\:{same}\:{line}\:...{since} \\ $$$$\angle{POQ}\:{is}\:=\mathrm{180}^{{o}} \\ $$$$ \\ $$