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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  (y/x)=((AE)/(EF))=((EB)/(EF))=(2/1)=2 ✓

$${set}\:{F}={midpoint}\:{of}\:{EB} \\ $$$${ED}//{FC} \\ $$$$\frac{{y}}{{x}}=\frac{{AE}}{{EF}}=\frac{{EB}}{{EF}}=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2}\:\checkmark \\ $$

Answered by deleteduser1 last updated on 04/Nov/23

Let DE and CB meet at F;join AF; in triangle  AFB; ⇒Line AC and FE are medians  ⇒D is the centroid⇒((AD)/(DC))=2

$${Let}\:{DE}\:{and}\:{CB}\:{meet}\:{at}\:{F};{join}\:{AF};\:{in}\:{triangle} \\ $$$${AFB};\:\Rightarrow{Line}\:{AC}\:{and}\:{FE}\:{are}\:{medians} \\ $$$$\Rightarrow{D}\:{is}\:{the}\:{centroid}\Rightarrow\frac{{AD}}{{DC}}=\mathrm{2} \\ $$

Answered by mr W last updated on 04/Nov/23

Commented by mr W last updated on 04/Nov/23

(y/x)=((AF)/(EC))=((AB)/(EB))=(2/1)=2 ✓

$$\frac{{y}}{{x}}=\frac{{AF}}{{EC}}=\frac{{AB}}{{EB}}=\frac{\mathrm{2}}{\mathrm{1}}=\mathrm{2}\:\checkmark \\ $$

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