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

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 \dots

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 \dots

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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