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Why Must the Angles Be Equal at Each Intersection?
At this point we know we are working with tilings of congruent triangles. So there are at most three different angles. For the sake of contradiction we'll approach this argument by looking at an intersection that could have all three angles present. After establishing a contradiction, we'll move on to only two angles and show that they must be equal.
Three different angles at an intersection
Consider the case of three different angles at an intersection. One of these angles is 90 degrees since our triangles are all right triangles. Label the other two α and β. The sum of the angles of a spherical triangle is greater than 180 degrees. So 90 + α + β >180. Or alternately α + β > 90. Now at any vertex with all three angles present they occur opposite and pair wise so that you get nα + mβ = 90.
But α + β > 90, so either n or m must be zero. Thus, at each vertex there can be at most two different angles.
Two different angles at an intersection
Now, consider a vertex with only two angles. Again, label them α and β. If one of them is the right angle of our triangle, then α + β > 90. If neither of them is our 90° angle then α + β is still greater than 90 because the sum of the angles in a spherical triangle is greater than 180.
You must have an α next to a β at the intersection. Then the remainder of the angles that occur up to the opposite α will be a number of α (n) and β (m). Now nα + mβ must be supplementary to α + β. So, you have α + β + nα + mβ = 180.
But, since α + β > 90, the above equation is only true if n or m is zero. Thus, at a vertex with two different angles, one of them must occur at most one time on one half of the vertex.
So our vertex looks like the one to the left.
Consider the triangle with α at a vertex. It has sides a, b and c opposite angles α, β and γ respectively. In particular b and c are the legs of α, and a and c are the legs of β.
Now, let's reconsider the intersection with α and β. Label the legs of the angles around the intersection, starting at one of the α angles.
Notice that when you do that one of the β angles must be bounded by side b. That can only happen if b = a. That means our original triangle is isosceles and implies α = β.
So, α must equal β and there are no intersections with two different angles.
When we assume there are three different angles at the intersection we find a contradiction. So we assume that there are only two different angles at the intersection. This assumption leads to the conclusion that those angles must be equal. Thus, at any intersection there is only one size angle present.