# $\frac{AM}{MB}=\frac{AN}{NC}$ iff $\frac{MQ}{QN}=1$

I have a triangle ABC and point $M \in AB$ and $N \in AC$. Defining the points $O= BN\cap CM$ and $Q=AO\cap MN$. I want to show that $\Large\frac {AM}{MB}=\frac {AN}{NC}$ iff $\Large\frac{MQ}{QN}=1$.

for $\Rightarrow$ I tried using Ceva's theorem and using the fact that $\Large \frac{MQ}{QN}=\frac{AA'}{A'B}$ (please tell me if this is wrong) where A' is the intersection of the line AO with BC.

Also $\Leftarrow$ I have no idea. Please help

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For forward direction, use Ceva's theorem on the triangle $AMN$ with point $O$, and the result falls our directly. – Calvin Lin Jan 29 '13 at 8:26

Applying Ceva's theorem to triangle $AMN$ with point $O$, we get that
$$\frac {AB}{BM} \frac {MQ}{QN} \frac {NC}{CA} = 1.$$
Hence, if $\frac {AM}{MB} = \frac {AN}{NC}$, then $\frac {AM+MB}{MB} = \frac {AN+NC}{NC}$, so $\frac {AB}{BM} \times \frac {NC}{CA} = 1$, which gives that $\frac {MQ}{QN} = 1$.
Conversely, if $\frac {MQ}{QN} = 1$, then $\frac {AB}{BM} = \frac {AC}{CN}$, and thus $\frac {AB-BM}{BM} = \frac {AC-CN}{CN}$, or that $\frac {AM}{BM} = \frac {AN}{CN}$.