If the median AD of the triangle ABC is bisec | vedclass

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(B) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Since $AD$ is the median,$D$ is the midpoint of $BC$,so $D = (\f

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$1: 3$

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(B) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Since $AD$ is the median,$D$ is the midpoint of $BC$,so $D = (\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2})$.$E$ is the midpoint of $AD$,so $E = (\frac{x_1 + \frac{x_2+x_3}{2}}{2}, \frac{y_1 + \frac{y_2+y_3}{2}}{2}) = (\frac{2x_1+x_2+x_3}{4}, \frac{2y_1+y_2+y_3}{4})$.Using the property of the centroid $G$ of triangle $ABC$,$G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.By applying Menelaus' theorem to triangle $ADC$ with transversal $B-E-F$,we find that $F$ divides $AC$ in the ratio $1: 2$.Thus,$AF: FC = 1: 2$,which implies $AF: AC = 1: (1+2) = 1: 3$.

If the median $AD$ of the triangle $ABC$ is bisected at $E$ and $BE$ meets $AC$ in $F$,then $AF: AC=$

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