In the diagram, $K L$ is parallel to $M N, A B=B C$, and $\angle K A C=50^{\circ}$. The value of $x$ is
(A) 40
(D) 100
(B) 65
(E) 80
(C) 25

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-133.jpg?height=317&width=504&top_left_y=1468&top_left_x=1277)
In the diagram, $K L$ is parallel to $M N, A B=B C$, and $\angle K A C=50^{\circ}$. The value of $x$ is
(A) 40
(B) 65
(C) 25
(D) 100
(E) 80

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-240.jpg?height=338&width=553&top_left_y=785&top_left_x=1339)

## Solution

Since $K L$ is parallel to $M N, \angle A C B$ and $\angle K A C$ are alternate angles. Thus, $\angle A C B=50^{\circ}$.

We are given that $\triangle B C A$ is isosceles, so $\angle B C A=\angle B A C=50^{\circ}$. We know that since angles $K A C$, $B A C$ and $L A B$ have a sum of $180^{\circ}$,

$$
\begin{aligned}
50^{\circ}+50^{\circ}+x^{\circ} & =180^{\circ} \\
x & =80
\end{aligned}
$$

ANSWER: (E)