Interesting problem! In order to solve this one, we’ll know some information about the measure of these triangles’ angles. First, remember that triangles have the very unique property that *equal angles sweep out equal lengths*. The other side of that is that if two sides of a triangle are congruent, we know the angles opposite them are congruent, too.
Since all the sides of triangle ABE are congruent, all its angles are congruent, and since the angles of a triangle add to 180, this means that each of the angles in ABE must be 60 degrees (since 60 + 60 + 60 = 180).
The other triangle of notice in the problem is triangle ABC. Notice that, since it’s formed from a square, it’s a *right* triangle. This means that one of its angles must be 90. Sharing two sides with the square ABCD also means that two of the triangle’s sides are congruent, which *also* means the two non-right angles must be congruent. This forces them to be 45 degrees (since 90 + 45 + 45 = 180).
Now, take a look at angle EAC. That small angle is what we end up with if we take angle EAB (an angle of triangle ABE) and subtract angle CAB (part of triangle ABC) from it. EAB measures 60 degrees, and CAB measures 45, so angle EAB must be 60 - 45 = 15 degrees.
At this point, we’re almost done. I want to bring your attention to the point where line segments EB and AC intersect; let’s call that point F. If we want to find the measure of x, we can look at the angle right across from it, angle AFE. Notice that this angle is one of three in the triangle AFE. We’ve already found the other two, actually: angle AEF is 60 degrees, and angle EAF is 15, so to find AFE, we can simply subtract those two from 180 to find a measure of
180 - 60 - 15 = 180 - 75 = 105 degrees.
And since AFE and x are vertical angles, we know x must also be 105 degrees.
