What is the angle at the midpoint of the line created by the intersection of two interior and two exterior angle bisectors of a triangle? The background for this question is that I occasionally tutor in various subjects and recently had a student ask me about this geometry question:

Let   I   and   E   be   the   intersection   points   of,   respectively,   the   interior   angle   bisectors   and   the   exterior   angle   bisectors   at   vertices   A   and   B   of   triangle   ABC.   If   <ABC   =   70'   and   M   is   the   midpoint   of   IE,   what   is   the   measure   of   <AMB?

He said that there wasn't any diagram or additional information included so I went ahead and made a sketch of what I believe the problem is describing:

I wasn't able to solve it using any interior/exterior angle bisector theorems and the only given information of <ABC = 70 wasn't enough for me to solve triangle ABC let alone find the measure of <AMB.
Is this a solvable problem given the information? What am I missing here?
 A: The problem may have a typo. The angles $\angle ACB$ and $\angle AMB$ are supplementary, which I will prove below. Thus if the specified angle is $\angle ACB$, then we can say that $\angle AMB = 110^\circ$. On the other hand, this means that we can always construct a different triangle where $\angle AMB$ doesn't change but $\angle ABC$ changes, by fixing $A$ and $B$ and moving $C$ along the circumcircle. So if only $\angle ABC$ is specified, the problem is not solvable.

It turns out that the midpoint of the line segment connecting the incenter and an excenter of a triangle lies on circumscribed circle. This is a consequence of the incenter-excenter lemma.
Namely, in the following diagram, let $I$ be the incenter, $I_A$ be the excenter opposite $A$, and $D$ be the midpoint of the arc $BC$ of the circumcircle, then $D$ is the center of a circle through $B$, $I$, $C$, $I_A$. (Watch out for the difference in notations! Your $A$, $B$, $C$, $E$, $M$ are $B$, $C$, $A$, $I_A$, $D$ here, respectively.)

The proof of the lemma is essentially angle chasing. The Evan Chen notes provide the proof that $BD = ID$. The same argument can be easily modified to show that $CD = ID$. I will show that $BD = I_AD$, which will complete the proof.
Note that $\angle I_A B D = \angle I_A B C - \angle DBC$. Note $\angle I_A BC$ is half the external angle at $B$ since $I_A$ is the excenter, so it equals $90^\circ - \frac{1}{2}\angle ABC$. Note also that $\angle DBC = \angle DAC$ since they are both subtended by arc $DC$, so $\angle DBC = \frac{1}{2}\angle BAC$. Combining the value of the two angles, we get $$\angle I_A B D = 90^\circ - \frac{1}{2}(\angle ABC + \angle BAC)\ .$$
On the other hand,
$$\begin{multline}\angle B I_A D = 180^\circ - \angle I_A BA - \angle BA I_A = 180^\circ - (\angle I_A BC + \angle ABC) - \angle BA I_A \\ = 180^\circ - \left(90^\circ - \frac{1}{2}\angle ABC + \angle ABC\right) - \frac{1}{2}\angle BAC = 90^\circ - \frac{1}{2}(\angle ABC + \angle BAC)\ . \end{multline}$$
Thus $\triangle BDI_A$ is isosceles, and $BD = I_A D$.

Now that we know $M$ is on the circumcircle, it's easy to see that $\angle AMB$ and $\angle ACB$ are subtended by opposite arcs, so they must be supplementary.
A: You are right. The measure of $\angle AMB$ isn't solvable from just the measure of $\angle ABC$. Here's why:
Points $A$, $B$, $I$ and $E$ all lie on the circle with diameter $IE$ and thereafter $\angle AMB=2\angle AEB=2\left(180-\angle AIB\right)=180-\angle C$.
Thus it's only the angle $\angle C$ which determines the value of $\angle AMB$. It was probably intended to mention $\angle ACB$ for which the problem has a solution.
