We have a $\triangle ABC,AB=6,BC=4, AC=5$. A line that is parallel to $AB$ intersects $AC$ and $BC$ at $M,N$, respectively. If $P_{\triangle CMN}=P_{ABNM}$, I should find the length of $MN$ ($P_A$ is the perimeter of figure $A$).
We have $\triangle ABC \sim \triangle MNC$ because $\angle BAC=\angle CMN$ and $\angle ACB=\angle MCN$. Therefore, $\dfrac{AB}{MN}=\dfrac{BC}{CN}=\dfrac{AC}{CM}$. Also we have $CM+MN+CN=AB+AM+BN+MN$ or $CM+CN=AB+AM+BN$. How to approach the problem further?
$$\dfrac{AB}{MN}=\dfrac{BC}{CN}=\dfrac{AC}{CM}=\dfrac 1f$$ Then use $CM=fAC=5f$, $CN=fBC=4f$, $MN=fAB=6f$. You also know $AM=AC-CM=5(1-f)$, and similarly for $BN=4(1-f)$. Just plug these into your last equation and find $f$.