Answer: all kids get the same amount.

Let $S$ be the area of the initial triangle, $S_A$ be the area of the quadrilateral piece, $S_B$ be the area of Berta's triangle, and $S_1$ and $S_2$ be the areas of the remaining two triangles (Fig. 14). Then $S_1 + S_B = S_2 + S_B = \frac{1}{2}S$ (a common altitude while the ratio of the corresponding bases being $\frac{1}{2}$) and $S_B = 2S_1$ (for similar reasons). Thus $S_1 = S_2 = \frac{1}{6}S$ and $S_B = \frac{1}{3}S$, whence also $S_1 + S_2 = \frac{1}{3}S$ and $S_A = \frac{1}{3}S$.

Suppose that mother makes one more cut from the third vertex to the midpoint of the opposite side. As all medians of a triangle meet in one point, the new cut divides Anna's and Berta's pieces into two parts while not touching Clara's pieces. So every child gets exactly two pieces. We show that medians of a triangle divide the triangle into six parts of equal area; this implies that all kids get the same amount of cake. Let the triangle be $ABC$, its medians be $AD, BE$ and $CF$, and the centroid be $G$ (Fig. 15). The length of the side $BD$ of the triangle $BGD$ is $\frac{1}{2}$ of the length of the side $BC$ of the triangle $ABC$, the length of the corresponding altitude in the triangle $BGD$ is $\frac{1}{3}$ of the length of the corresponding altitude in the triangle $ABC$ (since $|AD| = 3|GD|$, the perpendicular drawn from the point $A$ to the line $BC$ is 3 times longer than the perpendicular drawn from the point $G$ to the same line). Thus the area of the triangle $BGD$ equals $\frac{1}{6}$ of the area of the triangle $ABC$. The same holds for other pieces. Consequently, all pieces have the same area.