Lemma. Let $I$ be the incenter of $\triangle A B C$ and let the points $P$ and $Q$ lie on the lines $A B$ and $A C$. Then the points $A, I, P$, and $Q$ lie on a circle if and only if

$$\overline{B P}+\overline{C Q}=B C$$

where $\overline{B P}$ equals $|B P|$ if $P$ lies in the ray $B A \rightarrow$ and $-|B P|$ if it does not, and similarly for $\overline{C Q}$.
Proof of the lemma. We shall only consider the case when $P$ and $Q$ lie in the segments $A B$ and $A C$. All other cases are treated analogously.
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Suppose that $A, I, P$, and $Q$ lie on a circle. Let $D$ and $E$ be the contact points of the incircle of $\triangle A B C$ with $A B$ and $A C$. We have that $\angle P I Q=180^{\circ}-\alpha$, so $\angle D I P=\angle E I Q$ and, therefore, $\triangle D I P \simeq \triangle E I Q$. This gives us $D P=E Q$ and $B P+C Q=B D+C E=B C$, as needed.

The converse is established by following the foregoing chain of inequalities in reverse.

Let the circumcircle of $\triangle M I_{1} I_{3}$ meet the lines $A B, C M$, and $D M$ for the second time at $P, Q$, and $R$. By the lemma, $\overline{B P}+\overline{C Q}=B C$ and $\overline{D R}+\overline{A P}=D A$. Therefore, $\overline{C Q}+\overline{D R}=B C+D A-\overline{B P}-\overline{A P}=$ $B C+D A-A B$. Since $A B C D$ is circumscribed, this is equal to $C D$, and, by the lemma, the proof is complete.
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Second solution. Let $\omega_{1}, \omega_{2}$, and $\omega_{3}$ be the incircles of $\triangle M B C, \triangle M C D$, and $\triangle M D A$.
The common internal tangent $t_{1}$ of $\omega_{1}$ and $\omega_{2}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{2}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{1}\right]=\frac{1}{2}(M C+M D-C D-M B-M C+B C)$.
Analogously, the common internal tangent $t_{2}$ of $\omega_{2}$ and $\omega_{3}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{2}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{3}\right]=\frac{1}{2}(M C+M D-C D-M D-M A+D A)$.
Finally, the common external tangent $t_{3}$ of $\omega_{1}$ and $\omega_{3}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{1}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{3}\right]=\frac{1}{2}(M B+M C-B C+M D+M A-D A)$.
Since $A B C D$ is circumscribed, we have $A B+C D=B C+D A$, and, therefore, $t_{1}+t_{2}=t_{3}$. It follows from this that $\omega_{1}, \omega_{2}$, and $\omega_{3}$ have a common tangent $s$ (which separates $\omega_{2}$ from $\omega_{1}$ and $\omega_{3}$ ).
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Let $\triangle M K L$ be the triangle formed by the lines $M C, M D$, and $s$. Then, since $I_{1} I_{2}$ and $I_{2} I_{3}$ are external angle bisectors in it, we have $\angle I_{1} I_{2} I_{3}=90^{\circ}-\frac{1}{2} \angle K M L=180^{\circ}-\angle I_{1} M I_{3}$ and, therefore, $M I_{1} I_{2} I_{3}$ is cyclic.
$\operatorname{Remark}(\mathbf{P S C}):$ In the first solution, the angle $\alpha$ refers to the angle $\angle B A C$.
$\operatorname{Remark}(\mathbf{P S C})$ : Referring to solution 1 , the inequality $t_{3} \leq t_{1}+t_{2}$ holds, the equality is possible if and only if the circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ have a common tangent.

Remark(PSC): This problem was selected as G1 relative to the other Geometry problems proposed. PSC thinks the difficulty level of this problem is medium.