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Question-201373




Question Number 201373 by sonukgindia last updated on 05/Dec/23
Answered by aleks041103 last updated on 05/Dec/23
Commented by aleks041103 last updated on 05/Dec/23
AQ.BQ=DQ.FQ  BQ.CQ=FQ.EQ  ⇒((AQ)/(CQ))=((DQ)/(EQ))  ⇒orange:purple=red:blue  ⇒(orange=purple ⇔ red=blue)
$${AQ}.{BQ}={DQ}.{FQ} \\ $$$${BQ}.{CQ}={FQ}.{EQ} \\ $$$$\Rightarrow\frac{{AQ}}{{CQ}}=\frac{{DQ}}{{EQ}} \\ $$$$\Rightarrow{orange}:{purple}={red}:{blue} \\ $$$$\Rightarrow\left({orange}={purple}\:\Leftrightarrow\:{red}={blue}\right) \\ $$
Commented by aleks041103 last updated on 05/Dec/23
Commented by aleks041103 last updated on 05/Dec/23
△CFD≅△EFB   1. ∠CFD=∠EFB   2. ∠DCF=(1/2)DB^(⌢) =∠FEB   3.⇒∠CDF=∠EBF    ⇒((CF)/(EF))=((DF)/(BF))⇒BF.CF=DF.EF
$$\bigtriangleup{CFD}\cong\bigtriangleup{EFB} \\ $$$$\:\mathrm{1}.\:\angle{CFD}=\angle{EFB} \\ $$$$\:\mathrm{2}.\:\angle{DCF}=\frac{\mathrm{1}}{\mathrm{2}}\overset{\frown} {{DB}}=\angle{FEB} \\ $$$$\:\mathrm{3}.\Rightarrow\angle{CDF}=\angle{EBF} \\ $$$$ \\ $$$$\Rightarrow\frac{{CF}}{{EF}}=\frac{{DF}}{{BF}}\Rightarrow{BF}.{CF}={DF}.{EF} \\ $$
Commented by aleks041103 last updated on 05/Dec/23
Commented by aleks041103 last updated on 05/Dec/23
△BCF ≅ △BDE   1. ∠FBC=∠EBD   2. ∠BCF=(1/2)EF^(⌢)  =∠BDE   3. ∠BFC = ∠BED    ⇒((BF)/(BC))=((BE)/(BD))⇒BD.BF=BC.BE
$$\bigtriangleup{BCF}\:\cong\:\bigtriangleup{BDE} \\ $$$$\:\mathrm{1}.\:\angle{FBC}=\angle{EBD} \\ $$$$\:\mathrm{2}.\:\angle{BCF}=\frac{\mathrm{1}}{\mathrm{2}}\overset{\frown} {{EF}}\:=\angle{BDE} \\ $$$$\:\mathrm{3}.\:\angle{BFC}\:=\:\angle{BED} \\ $$$$ \\ $$$$\Rightarrow\frac{{BF}}{{BC}}=\frac{{BE}}{{BD}}\Rightarrow{BD}.{BF}={BC}.{BE} \\ $$
Answered by mr W last updated on 05/Dec/23
Commented by mr W last updated on 05/Dec/23
for big circle:  (b/(y+z))=(x/c)   ...(i)  for small circle:  y(y+z)=b(b+a)   ...(ii)  (i)×(ii):  y=((x(b+a))/c)  since b+a=c,  ⇒y=x ✓
$${for}\:{big}\:{circle}: \\ $$$$\frac{{b}}{{y}+{z}}=\frac{{x}}{{c}}\:\:\:…\left({i}\right) \\ $$$${for}\:{small}\:{circle}: \\ $$$${y}\left({y}+{z}\right)={b}\left({b}+{a}\right)\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$${y}=\frac{{x}\left({b}+{a}\right)}{{c}} \\ $$$${since}\:{b}+{a}={c}, \\ $$$$\Rightarrow{y}={x}\:\checkmark \\ $$

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