Question Number 208327 by efronzo1 last updated on 12/Jun/24
Answered by A5T last updated on 12/Jun/24
$${DE}=\mathrm{5}{x},{DF}=\mathrm{12}{x} \\ $$$${Let}\:{the}\:{perpendicular}\:{from}\:{A}\:{to}\:{BC}\:{meet}\:{it}\:{at}\:{H}; \\ $$$${AH}×{BC}=\mathrm{3}×\mathrm{4}=\mathrm{12}\Rightarrow{AH}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\Rightarrow{BH}=\sqrt{\mathrm{9}−\frac{\mathrm{144}}{\mathrm{25}}}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\frac{{AD}}{{AB}}=\frac{{DE}}{{BC}}={x}\Rightarrow{AD}=\mathrm{3}{x}\Rightarrow{AE}=\mathrm{4}{x} \\ $$$${BF}×{BC}={BD}×{BA}\Rightarrow{BF}=\frac{\mathrm{3}×\left(\mathrm{3}−\mathrm{3}{x}\right)}{\mathrm{5}}…\left({i}\right) \\ $$$$\frac{{BF}}{{BH}}=\frac{{DF}}{{AH}}\Rightarrow{BF}=\frac{\frac{\mathrm{9}×\mathrm{12}{x}}{\mathrm{5}}}{\frac{\mathrm{12}}{\mathrm{5}}}=\mathrm{9}{x}…\left({ii}\right) \\ $$$$\left({i}\right)=\left({ii}\right)\Rightarrow\mathrm{3}−\mathrm{3}{x}=\mathrm{15}{x}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{6}}\Rightarrow{DF}=\mathrm{12}{x}=\mathrm{2} \\ $$