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Question-42112




Question Number 42112 by Raj Singh last updated on 18/Aug/18
Commented by Raj Singh last updated on 18/Aug/18
if OF and OE is bisector of <BOC  and <AOC and OE perpendicalr  to OF  then prove that A O B is linear
$${if}\:{OF}\:{and}\:{OE}\:{is}\:{bisector}\:{of}\:<{BOC} \\ $$$${and}\:<{AOC}\:{and}\:{OE}\:{perpendicalr} \\ $$$${to}\:{OF}\:\:{then}\:{prove}\:{that}\:{A}\:{O}\:{B}\:{is}\:{linear} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Aug/18
oF bisector of∠BOC  ∠BOF=∠FOC=x  ∠BOC=2x  ∠AOE=∠EOC=y  ∠AOC=2y  Given EO⊥OF   so  ∠EOC+∠COF=y+x=90^o   now ∠AOC+∠BOC=2x+2y=2(x+y)=180^o   hence A,O andB lies on a line
$${oF}\:{bisector}\:{of}\angle{BOC} \\ $$$$\angle{BOF}=\angle{FOC}={x} \\ $$$$\angle{BOC}=\mathrm{2}{x} \\ $$$$\angle{AOE}=\angle{EOC}={y} \\ $$$$\angle{AOC}=\mathrm{2}{y} \\ $$$${Given}\:{EO}\bot{OF}\:\:\:{so} \\ $$$$\angle{EOC}+\angle{COF}={y}+{x}=\mathrm{90}^{{o}} \\ $$$${now}\:\angle{AOC}+\angle{BOC}=\mathrm{2}{x}+\mathrm{2}{y}=\mathrm{2}\left({x}+{y}\right)=\mathrm{180}^{{o}} \\ $$$${hence}\:{A},{O}\:{andB}\:{lies}\:{on}\:{a}\:{line} \\ $$

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