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_ac36362783317e0251fdg-117.jpg?height=307&width=376&top_left_y=2042&top_left_x=1384)
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_33a8a6559d675f59baa2g-230.jpg?height=363&width=447&top_left_y=827&top_left_x=1403)

Answer: (E)

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