, 2021

Solution:

We first note that the circumcircles of $D B Z$ and $Y D C$ are tangent to $B C$ from our angle criteria. By power of a point, we obtain that $M$ lies on the radical axis of the two circles and clearly $D$ does as well. Therefore, we find that $A$ lies on the radical axis so $A Y \cdot A C = A B \cdot A Z$ implying that $B Y Z C$ is a cyclic quadrilateral.

Next, by Reim's Theorem on $(B Y Z C)$ and $(D Y Z)$, we get that $(D Y Z)$ intersects $A B, A C$ at $B^{\prime}, C^{\prime}$ where $B C, B^{\prime} C^{\prime}$ are parallel. Then a negative homothety maps $B$ to $B^{\prime}$ and $C$ to $C^{\prime}$, so $(D B C)$ gets mapped to $\left(D B^{\prime} C^{\prime}\right)$, and we have tangent circles.

Solution:

Let $(D Y Z)$ intersect $A B$ and $A C$ at $B^{\prime}$ and $C^{\prime}$, respectively. We see that $\measuredangle Y C^{\prime} B^{\prime} = \measuredangle Y Z B^{\prime} = \measuredangle Y Z B = \measuredangle Y C B$. Thus, $B C \parallel B^{\prime} C^{\prime}$. This means that there exists a negative homothety taking $B$ to $B^{\prime}$ and $C$ to $C^{\prime}$ which will map $(D B C)$ to $\left(D B^{\prime} C^{\prime}\right)$ which is also $(D Y Z)$.