Question

what are the values of x and y?

Explanation:

Step1: Use the SAS (Side - Angle - Side) congruence criterion

Since two sides are equal (marked with the same tick marks) and the included angles (the angles between the equal sides) are considered. By the SAS congruence, \(\triangle DEF\cong\triangle FED\) (assuming the sides and the included parts). But more straightforwardly, for the given triangle - like figure (assuming it's a congruent triangle situation where we can use triangle properties). For the angle \(x\):
We know that in a triangle (if we assume the two triangles formed by the equal sides are congruent), the angle \(x\) is equal to the angle opposite to the side with the same length relation. Using the property that in congruent triangles (by SAS as two sides and the included side (if we consider the side common or the equal - side structure) the corresponding angles are equal. If we assume a triangle where one angle is \(31^{\circ}\), and using the property that the sum of angles in a triangle is \(180^{\circ}\) is not directly needed here. Wait, actually, if we consider the two triangles (from the equal - side markings) are congruent. For angle \(x\):
Since the side - side equality (marked), and if we assume the triangles are congruent (by SSS or SAS, but with the given markings, assume SAS - like congruence for angle equality). The angle \(x = 31^{\circ}\) (corresponding angles of congruent triangles).

Step2: Use the angle - sum property of a triangle (if needed for \(y\), but actually, if we consider the two triangles (from the side - markings) and the fact that in an isosceles - like (from congruence) situation.

Another approach: Let's assume the two triangles (divided by the common side \(EF\) (if we consider the figure as two triangles with \(ED = FD\) (from the tick - mark) and \(EF\) common (or another side equality). Wait, no, re - evaluating. If we consider the two triangles (from the side - markings \(ED = FD\) (one tick) and \(EF\) is common (assuming the figure is two triangles \(\triangle EFD\) and \(\triangle DEF\) - no, better to use the property of congruent triangles. Since \(ED = FD\) (side - side equality from tick - marks), and \(EF\) is common (if we assume the figure is two triangles \(\triangle EFD\) and \(\triangle DEF\) with \(ED = FD\), \(EF=EF\) (common side), and then by SSS (if the third side is equal, but no, wait, no - the problem is about angles. Wait, actually, if we consider the triangle \(\triangle EFD\) where \(ED = FD\) (so it's an isosceles triangle). Wait, no, the markings: one tick on \(ED\) and \(FD\) (assuming \(ED = FD\)), and then \(\angle EFD = 31^{\circ}\). Then, if we consider the congruence (maybe \(\triangle EFD\) and \(\triangle DEF\) - no, better:
Let’s use the property of congruent triangles (SAS: if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle). But here, maybe it's a mis - drawn figure. Wait, no, another approach:
If we assume that the two triangles (from the side - markings) are congruent. For \(x\): since \(\angle EFD = 31^{\circ}\), and by congruence (corresponding angles), \(x = 31^{\circ}\). For \(y\):
We know that the sum of angles in a triangle is \(180^{\circ}\). If we assume a triangle (say \(\triangle EFD\)) with two angles: one is \(31^{\circ}\) (\(\angle EFD\)), another is \(x = 31^{\circ}\) (if it's a congruent - triangle - derived angle, no, wait, no. Wait, re - doing:
Let’s consider the two triangles (from the side - markings \(ED = FD\) (one tick), and assume \(EF\) is a common side. Then \(\triangle EFD\cong\triangle DEF\) (by SSS: \(ED = FD\), \(EF = EF\), and assume \(DF = ED\) - no, wait, no, the problem is about angles. Wait, actually, if we consider the triangle \(\triangle EFD\) where \(ED = FD\) (so it's an isosceles triangle). Wait, no, the markings: one tick on \(ED\) and \(FD\) (so \(ED = FD\)), and \(\angle EFD = 31^{\circ}\). Then, using the property of isosceles triangles (if \(ED = FD\), then \(\angle DEF=\angle DFE = 31^{\circ}\) (no, wait, no - wrong. Wait, no, re - think.
Wait, the correct approach:
Since \(ED = FD\) (side - side equality from tick - marks) and assume \(EF\) is a common side. Then \(\triangle EFD\cong\triangle DEF\) (by SSS: \(ED = FD\), \(EF = EF\), \(DF = ED\) (from tick - mark). Then, \(\angle EDF=\angle DFE = 31^{\circ}\) (corresponding angles of congruent triangles for \(x\)). For \(y\):
We know that in \(\triangle EFD\), \(\angle FED + \angle EDF+\angle DFE=180^{\circ}\). But since \(\triangle EFD\cong\triangle DEF\), \(\angle FED=\angle EDF\) (if \(ED = FD\) and \(EF\) common). Wait, no, better:
Another way: Let’s assume that the two triangles (from the side - markings) give us that \(x = 31^{\circ}\) (corresponding angles). And for \(y\):
Since the sum of angles in a triangle (if we consider a triangle with angles \(y\), \(x\), and \(31^{\circ}\) - no, wait, no. Wait, actually, if we consider the two triangles (from congruence) \(\triangle EFD\) and \(\triangle DEF\). \(\angle EDF=x = 31^{\circ}\) (corresponding to \(\angle DFE\)). And for \(y\):
In \(\triangle EFD\), if we assume it's a triangle with \(\angle EDF = 31^{\circ}\), \(\angle DFE = 31^{\circ}\), then \(y=180-(31 + 31)=118^{\circ}\) (sum of angles in a triangle \(=180^{\circ}\)). But no, wait, no - recheck.
Wait, no, the correct approach:
Since \(ED = FD\) (side - side equality), and assume \(EF\) is a common side. Then \(\triangle EFD\cong\triangle DEF\) (SSS). Then \(\angle EDF=\angle DFE = 31^{\circ}\) (so \(x = 31^{\circ}\)). And \(\angle FED=\angle EFD\) (but no, wait, no - wait, in \(\triangle EFD\), \(\angle EDF=x\), \(\angle DFE = 31^{\circ}\), and \(\angle FED=y\). Using the sum of angles in a triangle (\(\angle EDF+\angle DFE+\angle FED = 180^{\circ}\)), so \(x + 31+y=180\). But since \(x = 31\) (from congruence - corresponding angles, because \(ED = FD\) and \(EF\) is common (SSS congruence makes \(\angle EDF=\angle DFE\)). Then \(31+31 + y=180\), so \(y=180-(31 + 31)=118\).

Answer:

\(x = 31^{\circ}\), \(y = 118^{\circ}\)