Let $B = (0,0), E = (a,0), C = (a+b,0), F = (a+b+\frac{ac}{a+b},ta), D =(a+b+c, ta+tb), A = (c, ta+tb)$. It is easy to see that the condition is valid here. Now, line $BD$ has equation $y = \frac{t(a+b)}{a+b+c}x$ and line $AE$ has the equation $y = \frac{t(a+b)}{c-a} (x-a)$. We can solve this to get the x coordinate of $M$ to be $\frac{a(a+b+c)}{2a+b}$. Similarly, equation of line $AF$ is $y = \frac{tb(a+b)}{cb-(a+b)^2}(x-c) + t(a+b)$. We can solve it again to get x-coordinate of $N$ to be $\frac{(a+b)(a+b+c)}{a+2b}$. Now, we notice that the ratio of $BM : MN : ND$ is equal to $a^2+2ab:a^2+b^2+ab:2ab+b^2$, which is valid triangle since none of them is at least half of the sum of the two others.