In the diagram, $\triangle X Y Z$ is isosceles with $X Y=X Z$. Also, point $W$ is on $X Z$ so that $X W=W Y=Y Z$. The measure of $\angle X Y W$ is
(A) $18^{\circ}$
(B) $30^{\circ}$
(C) $45^{\circ}$
(D) $36^{\circ}$
(E) $60^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-076.jpg?height=444&width=341&top_left_y=1339&top_left_x=1293)
Let $\angle X Y W=\theta$.

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

Since the sum of the angles in $\triangle X Y W$ is $180^{\circ}$, then $\angle X W Y=180^{\circ}-2 \theta$

Since $\angle X W Y+\angle Z W Y=180^{\circ}$, then $\angle Z W Y=180^{\circ}-\left(180^{\circ}-2 \theta\right)=2 \theta$.

Since $\triangle Y W Z$ is isosceles with $Y W=Y Z$, then $\angle Y Z W=\angle Z W Y=2 \theta$.

Since $\triangle X Y Z$ is isosceles, with $X Y=X Z$, then $\angle X Y Z=\angle X Z Y=2 \theta$.

Since the sum of the angles in $\triangle X Y Z$ is $180^{\circ}$, then $\angle X Y Z+\angle X Z Y+\angle Y X Z=180^{\circ}$ or $2 \theta+2 \theta+\theta=180^{\circ}$, or $5 \theta=180^{\circ}$, or $\theta=36^{\circ}$.

ANSWER: (D)