Answer: C

Small Hint:

A segment parallel to the bases divides both legs in the same fraction

Big Hint:

Let the upper leg segments be $4t$ and $6t$; the common dividing segment cancels when the two perimeters are equated

Solution:

Let the upper pieces of the legs of lengths $4$ and $6$ be $4t$ and $6t,$ respectively. The dividing segment appears once in each perimeter and cancels. Equal perimeters therefore give $$\begin{aligned} 3+4t+6t &=9+4(1-t)\\ &\quad+6(1-t). \end{aligned}$$ Hence $20t=16,$ so $t=\frac{4}{5}.$ Each leg is divided in the ratio $$t:(1-t)=\frac45:\frac15=4:1.$$

Thus, the correct answer is C.