The answer is $90^\circ$. The base is partitioned into four segments in the ratio $x : x : y : 2x+y$. Suppose the length of the left-hand side of the triangle is 1. Then the length of the angle bisector is also 1. Applying the angle bisector theorem to the large triangle, we see that the length of the right-hand side is $\frac{2x+2y}{2x}= 1+ \frac{y}{x}$ . But if we apply the angle bisector theorem to the left half of the triangle, we obtain $\frac{2x+y}{y} = 1+ \frac{2x}{y}$ for the same length. Therefore $\frac{y}{x}= \frac{2x}{y}$ , so $x : y = 1 :\sqrt{2}$. Now apply the angle bisector theorem a third time to the right triangle formed by the altitude and the median. The segments in the base are in the ratio $x : y = 1 : \sqrt{2}$, so the altitude and the median form the same ratio. As this is a right triangle, it must be a $45-45-90$ triangle. So the quadrisected angle is right.