Question Number 70672 by ajfour last updated on 06/Oct/19
Commented by ajfour last updated on 06/Oct/19
$${Find}\:{sides}\:{of}\:\bigtriangleup{ABC}.\:{Also}\:{find} \\ $$$$\:{the}\:{same}\:{if}\:{instead}\:{of}\:{EF}\:, \\ $$$$\:{DF}=\mathrm{2}\:. \\ $$
Answered by mind is power last updated on 06/Oct/19
$${x}={AD} \\ $$$$\Rightarrow\frac{\mathrm{2}}{{x}}=\frac{{EB}}{{EB}+\mathrm{3}}\Rightarrow\mathrm{2}{EB}+\mathrm{6}={xEB}\Rightarrow{EB}=\frac{\mathrm{6}}{{x}−\mathrm{2}} \\ $$$$\Rightarrow\frac{{EA}}{{AB}}=\frac{{AD}}{{AC}}\Leftrightarrow\frac{\mathrm{3}}{\mathrm{3}+\frac{\mathrm{6}}{{x}−\mathrm{2}}}=\frac{{x}}{{x}+\mathrm{5}}\Rightarrow\frac{\left({x}−\mathrm{2}\right)}{{x}}=\frac{{x}}{{x}+\mathrm{5}}\Rightarrow{x}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{10} \\ $$$$\Rightarrow{x}=\frac{\mathrm{10}}{\mathrm{3}}={AD} \\ $$$${EB}=\frac{\mathrm{18}}{\mathrm{7}},{BF}=\sqrt{\left(\frac{\mathrm{18}}{\mathrm{7}}\right)^{\mathrm{2}} −\mathrm{4}}=\sqrt{\frac{\mathrm{128}}{\mathrm{49}}}=\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{7}} \\ $$$$\frac{{BF}}{{BD}}=\frac{\mathrm{2}}{{AD}}\Rightarrow{DB}=\frac{{BF}.{AD}}{\mathrm{2}}=\frac{\mathrm{40}\sqrt{\mathrm{2}}}{\mathrm{21}} \\ $$$${FD}=\frac{\mathrm{40}\sqrt{\mathrm{2}}}{\mathrm{21}}−\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{7}}=\frac{\mathrm{16}\sqrt{\mathrm{2}}}{\mathrm{7}} \\ $$
Commented by ajfour last updated on 06/Oct/19
$${excellent},\:{thanks}!\:{but}\:\:\:{EB}=\frac{\mathrm{18}}{\mathrm{4}}. \\ $$
Commented by mind is power last updated on 06/Oct/19
$${yeah} \\ $$