In the diagram, $B, C$ and $D$ lie on a straight line, with $\angle A C D=100^{\circ}$, $\angle A D B=x^{\circ}, \angle A B D=2 x^{\circ}$, and $\angle D A C=\angle B A C=y^{\circ}$. The value of $x$ is
(A) 10
(B) 45
(C) 30
(D) 50
(E) 20

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-118.jpg?height=360&width=434&top_left_y=119&top_left_x=1377)
In triangle $A C D, x^{\circ}+y^{\circ}+100^{\circ}=180^{\circ}$, so $x+y=80$

Since $\angle A C B$ and $\angle A C D$ are supplementary, then

$$
\angle A C B=180^{\circ}-\angle A C D=80^{\circ}
$$

Thus, in triangle $A C B, 2 x^{\circ}+y^{\circ}+80^{\circ}=180^{\circ}$, so $2 x+y=100 \quad(* *)$.

Subtracting $(*)$ from $(* *)$, we obtain $x=20$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-215.jpg?height=363&width=441&top_left_y=995&top_left_x=1384)

Answer: (E)