[ Similar triangles ]

In triangle $ABC$, with sides $a, b$ and $c$ given, a line $MN$ parallel to $AC$ is drawn such that $AM = BN$. Find $MN$.

#
Consider similar triangles.

## Solution

Let $A M=B N=x$ and find $x$ using the similarity of triangles $B M N$ and $B A C:{ }^{c-x /{ }_{c}}={ }^{x / a} \Rightarrow \quad x={ }^{a c} / a+c$. From the same similarity, we find that $M N={ }^{x b} / a=b c / a+c$.

## Answer

$b c / a+c$.