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

Dec 28, 2016

See below.

Explanation:

We will apply the theorem of Menelaus of Alexandria to the sub-triangle $\Delta E C B$ and the line $\left[A D\right]$

According to Menelaus,

$\frac{\left\mid A C \right\mid}{\left\mid A E \right\mid} \times \frac{\left\mid E F \right\mid}{\left\mid F B \right\mid} \times \frac{\left\mid B D \right\mid}{\left\mid D C \right\mid} = 1$

but $\left\mid A E \right\mid = \left\mid E F \right\mid$ and $\left\mid B D \right\mid = \left\mid D C \right\mid$ so we have

$\frac{\left\mid A C \right\mid}{\left\mid F B \right\mid} = 1$