The fact that $\triangle EMI$ is isosceles, namely $IM = EM$, is a direct consequence of the fact that $\angle EMI = 90^\circ$ and $M$ is the midpoint of $\overline{BC}$. To see this, all one needs to do is consider a rotation of the entire figure about point $M$ by $90$ degrees, either clockwise or counterclockwise. Doing so will identify $\triangle AMB$ with $\triangle AMC$; moreover, it also identifies $E$ with $I$, since $\angle EMI = 90^\circ$ as previously stated. Hence $IM = EM$ as claimed.
One simply needs to flesh out the argument a bit better. Moreover, the argument is not strictly complete because it only shows that if a point lies on such a path, it must satisfy the absolute value criterion, but it does not explicitly state that there exists a valid passing through a given point that satisfies the absolute value criterion.
To understand the motivation, consider an arbitrary lattice point $C = (x,y)$. We know that such a point must either belong to the set of "desired" points--i.e., there is a path from $A$ to $B$ of at length at most $20$ that passes through it--or, it does not. There clearly cannot be any points whose classification is "ambiguous/unknown."
Now if a point is in the desired set of points, the existence of a path from $A$ to $B$ through $C = (x,y)$ implies that its length is the sum of the lengths of the subpaths from $A$ to $C$ and then from $C$ to $B$. If use the notation $L(A,B)$ to denote this length, then $$20 \ge L(A,B) = L(A,C) + L(C,B) \ge 0.$$
Next, we observe that for any two lattice points $P= (x_1, y_1), Q = (x_2, y_2)$, the length of the shortest path along the lattice must be $|x_1 - x_2| + |y_1 - y_2|$. To see why, we reason that we must move a minimum distance $|x_1 - x_2|$ in the horizontal direction, and a minimum distance $|y_1 - y_2|$ in the vertical. So we deduce $$L(A,C) + L(C,B) \ge (|-3 - x| + |2 - y|) + (|x - 3| + |y - (-2)|).$$ Therefore, if $C$ is in the desired set of points, it satisfies the claimed absolute value inequality.
Conversely, it is easy to show that if $C$ satisfies the absolute value inequality, then a path exists from $A$ to $B$ through $C$ with length not more than $20$: simply choose a route that moves horizontally from $A = (-3,2)$ to $A' = (x,2)$, with length $|x+3|$, then vertically from $A' = (x,2)$ to $C = (x,y)$ with length $|y-2|$; then from $C$ to $B' = (3,y)$ with length $|3-x|$, and finally from $B'$ to $B = (3,-2)$ with length $|y+2|$, and since the sum of these lengths does not exceed $20$ (as we supposed that $C$ satisfies the absolute value inequality), this path meets the desired criteria.
Best Answer
Take the radius of the earth as 1. $A,C$ lie in the plane containing the equator. Project $B$ onto that plane to give a point $B'$. The angle $ACB'$ is $135^\circ$ ($=360^\circ-110^\circ-115^\circ$). $AC=1,CB'=\frac{1}{\sqrt2}$ (because $B$ is at latitude $45^\circ$). So by the cosine formula we find $AB'=\sqrt{\frac{5}{2}}$.
The triangle $AB'B$ has a right angle at $B'$. So the hypotenuse $AB=\sqrt3$.
Now the triangle $ABC$ is isosceles with $AB=\sqrt3$ and $AC=BC=1$. So $\sin\frac{1}{2}ACB=\frac{\sqrt3}{2}$ and hence $\frac{1}{2}ACB=60^\circ$ and so $\angle ACB=120^\circ$.