In $Δ A B C$ , $A D$ , is a median and $E$ is the mid point of $A D$ . If $B E$ is produced to meet $A C$ in $F$ , show that $A F = \frac{1}{3} A C$ .

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Solution

$A D$ is median $\Rightarrow B D = C D$

$E$ is mid point of $A D$

Draw a line $D G$ , such that $D G | | B F$

In $△ A D G$
$E F | | D G$ and $E$ is mid point of $A D$

$∴ F$ is midpoint of $A G$ [converse of midpoint theorem]

$\Rightarrow A F = F G - - - 1$

In $△ B C F$
$D$ is midpoint of $B C$ and $D G | | B F$

$∴ F G = G C - - - 2$

From $1$ and $2$

$∴ A F = F G = G C$

$A C = A F + F G + G C$

$\Rightarrow \frac{1}{3} A C = A F$

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E is the mid-point of a median AD of $Δ A B C$ and BE is produced to meet AC at F. Show that $A F = \frac{1}{3} A C$ .

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