The-median-AD-of-triangle-ABC-is-bisected-at-E-and-BE-meets-AC-at-F-Find-AF-FC-

Question Number 42180 by rahul 19 last updated on 19/Aug/18

The median AD of triangle ABC is

bisected at E and BE meets AC at F .

Find AF : FC .

Answered by MJS last updated on 19/Aug/18

you can put any triangle a b c in this position :

δ = {Heron}^{'} s Number = \sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}

A = (\begin{matrix} 0 \\ 0 \end{matrix}) B = (\begin{matrix} c \\ 0 \end{matrix}) C = (\begin{matrix} \frac{- a^{2} + b^{2} + c^{2}}{2 c} \\ \frac{δ}{2 c} \end{matrix})

D = \frac{B + C}{2} = (\begin{matrix} \frac{- a^{2} + b^{2} + 3 c^{2}}{4 c} \\ \frac{δ}{4 c} \end{matrix})

E = \frac{A + D}{2} = \frac{2 A + B + C}{4} = (\begin{matrix} \frac{- a^{2} + b^{2} + 3 c^{2}}{8 c} \\ \frac{δ}{8 c} \end{matrix})

l i n e B E : y = \frac{δ}{a^{2} - b^{2} + 5 c^{2}} (c - x)

l i n e A C : y = \frac{δ}{- a^{2} + b^{2} + c^{2}} x

F = B E \cap A C = (\begin{matrix} \frac{- a^{2} + b^{2} + c^{2}}{6 c} \\ \frac{δ}{6 c} \end{matrix})

∣ A F ∣= \frac{\sqrt{δ^{2} + {(- a^{2} + b^{2} + c^{2})}^{2}}}{6 c}

∣ F C ∣= \frac{\sqrt{δ^{2} + {(- a^{2} + b^{2} + c^{2})}^{2}}}{3 c}

\frac{∣ A F ∣}{∣ F C ∣} = \frac{1}{2}

Commented by rahul 19 last updated on 20/Aug/18

Sir, how u write coordinates of C ?

Commented by MJS last updated on 20/Aug/18

a = B C b = C A c = A B

I put A = point of origin and A B on x - axis \Rightarrow

\Rightarrow B = (\begin{matrix} c \\ 0 \end{matrix})

to find C we have 2 rectangular triangles

c = p + q \Rightarrow p^{2} + h_{c}^{2} = b^{2} \land q^{2} + h_{c}^{2} = a^{2}

C = (\begin{matrix} p \\ h_{c} \end{matrix})

we know that the area of A B C is

(1) \frac{{c h}_{c}}{2}

(2) \frac{δ}{4}

\frac{{c h}_{c}}{2} = \frac{δ}{4} \Rightarrow h_{c} = \frac{δ}{2 c}

p^{2} + h_{c}^{2} = b^{2}

p^{2} = b^{2} - h_{c}^{2}

p^{2} = \frac{4 b^{2} c^{2} - δ^{2}}{4 c^{2}}

[δ^{2} = 2 (a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2}) - (a^{4} + b^{4} + c^{4})]

p^{2} = \frac{a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2} + 2 b^{2} c^{2}}{4 c^{2}}

p^{2} = \frac{{(- a^{2} + b^{2} + c^{2})}^{2}}{4 c^{2}}

p = \frac{- a^{2} + b^{2} + c^{2}}{4 c^{2}}

C = (\begin{matrix} p \\ h_{c} \end{matrix}) = (\begin{matrix} \frac{- a^{2} + b^{2} + c^{2}}{4 c^{2}} \\ \frac{δ}{2 c} \end{matrix})

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