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