Question Number 199510 by universe last updated on 04/Nov/23
Answered by mr W last updated on 06/Nov/23
Commented by mr W last updated on 06/Nov/23
$${set}\:{F}={midpoint}\:{of}\:{EB} \\ $$$${ED}//{FC} \\ $$$$\frac{{y}}{{x}}=\frac{{AE}}{{EF}}=\frac{{EB}}{{EF}}=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2}\:\checkmark \\ $$
Answered by AST last updated on 04/Nov/23
$${Let}\:{DE}\:{and}\:{CB}\:{meet}\:{at}\:{F};{join}\:{AF};\:{in}\:{triangle} \\ $$$${AFB};\:\Rightarrow{Line}\:{AC}\:{and}\:{FE}\:{are}\:{medians} \\ $$$$\Rightarrow{D}\:{is}\:{the}\:{centroid}\Rightarrow\frac{{AD}}{{DC}}=\mathrm{2} \\ $$
Answered by mr W last updated on 04/Nov/23
Commented by mr W last updated on 04/Nov/23
$$\frac{{y}}{{x}}=\frac{{AF}}{{EC}}=\frac{{AB}}{{EB}}=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2}\:\checkmark \\ $$