Given $A B C$ $ABC$ a triangle where $¯ ¯¯¯¯ ¯ A D$ $bar(AD)$ is the median and let the segment line $¯ ¯¯¯¯ ¯ B E$ $bar(BE)$ which meets $¯ ¯¯¯¯ ¯ A D$ $bar(AD)$ at $F$ $F$ and $¯ ¯¯¯¯ ¯ A C$ $bar(AC)$ at $E$ $E$ . If we assume that $¯ ¯¯¯¯ ¯ A E = ¯ ¯¯¯¯ ¯ E F$ $bar(AE)=bar(EF)$ , show that $¯ ¯¯¯¯ ¯ A C = ¯ ¯¯¯¯ ¯ B F$ $bar(AC)=bar(BF)$ ?.

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Answer 1 · 2016-12-28T17:53:41

See below.

We will apply the theorem of Menelaus of Alexandria to the sub-triangle $Delta ECB$ and the line $[AD]$

According to Menelaus,

$abs(AC)/abs(AE) xx abs(EF)/abs(FB) xx abs(BD)/abs(DC)=1$

but $abs(AE)=abs(EF)$ and $abs(BD)=abs(DC)$ so we have

$abs(AC)/abs(FB)=1$

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