In the diagram, $M$ is the midpoint of $Y Z, \angle X M Z=30^{\circ}$, and $\angle X Y Z=15^{\circ}$. The measure of $\angle X Z Y$ is
(A) $75^{\circ}$
(B) $65^{\circ}$
(C) $60^{\circ}$
(D) $80^{\circ}$
(E) $85^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-040.jpg?height=225&width=504&top_left_y=1292&top_left_x=1163)
Since $\angle X M Z=30^{\circ}$, then $\angle X M Y=180^{\circ}-\angle X M Z=180^{\circ}-30^{\circ}=150^{\circ}$.

Since the angles in $\triangle X M Y$ add to $180^{\circ}$, then

$$
\angle Y X M=180^{\circ}-\angle X Y Z-\angle X M Y=180^{\circ}-15^{\circ}-150^{\circ}=15^{\circ}
$$

(Alternatively, since $\angle X M Z$ is an exterior angle of $\triangle X M Y$, then $\angle X M Z=\angle Y X M+\angle X Y M$ which also gives $\angle Y X M=15^{\circ}$.)

Since $\angle X Y M=\angle Y X M$, then $\triangle X M Y$ is isosceles with $M X=M Y$.

But $M$ is the midpoint of $Y Z$, and so $M Y=M Z$.

Since $M X=M Y$ and $M Y=M Z$, then $M X=M Z$.

This means that $\triangle X M Z$ is isosceles with $\angle X Z M=\angle Z X M$.

Therefore, $\angle X Z Y=\angle X Z M=\frac{1}{2}\left(180^{\circ}-\angle X M Z\right)=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$.

ANSWER: (A)