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


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Question Number 54422 by gunawan last updated on 03/Feb/19

$The median A D of the triangle A B C$ $is bisected at E, B E meets A C in F,$ $then A F : A C =$
Commented by ajfour last updated on 03/Feb/19

Commented by ajfour last updated on 03/Feb/19

$l e t B C = \bar{a}, B A = \bar{c}, A F = λ (\bar{a} - \bar{c})$ $e q . o f B F : {\bar{r}}_{F} = μ (\bar{c} + \frac{\bar{a}}{2}) = \bar{c} + λ (\bar{a} - \bar{c})$ $\Rightarrow μ = 1 - λ a n d \frac{μ}{2} = λ$ $\Rightarrow 2 λ = 1 - λ$ $h e n c e λ = \frac{1}{3} o r \frac{A F}{A C} = \frac{1}{3} .$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

$t a k i n g t h e h e l p o f d r a w i n g o f s i r A j f o u r$ $a l t e r n a t i v e p r o o f . . .$ $d r a w D K ∣∣ C F$ $\frac{B D}{D C} = \frac{B K}{K F} [s i n c e D K ∣∣ C F]$ $\frac{B D}{B C} = \frac{B K}{B F} = \frac{1}{2} s o K D = \frac{1}{2} F C$ $△ D K E a n d △ A E F$ $< D E K =< A E F [v e r t i c a l l y o p p o s i t e a n g l e]$ $A E = E D g i v e n$ $< K D E =< E A F [a l t e r n a t e a n g l e]$ $△ s a r e c o n g u r e n t . . .$ $s o K D = A F = x (a s p e r p i c t u r e)$ $A C = A F + F C = x + 2 x = 3 x$ $s o \frac{A F}{A C} = \frac{x}{3 x} = \frac{1}{3}$
	Answered by mr W last updated on 03/Feb/19

	Commented by ajfour last updated on 03/Feb/19

	$w o w! m a n t h i n k s, G o d l a u g h s!$
	Commented by mr W last updated on 03/Feb/19

	$d r a w D G / / B F$ $∵ A E = E D a n d E F / / D G$ $∴ A F = F G$ $∵ D G / / B F a n d B D = D C$ $∴ F G = G C$ $∵ A F = F G = G C$ $∴ A F / A C = 1 / 3$
	Commented by gunawan last updated on 05/Feb/19

	$thank you Sir$
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