Question

In Figure, find the values of $x$ and $y$ and then show that $AB\parallel CD$.

Answer 1

Step-1 Compute the values of $x$ and $y$

Given: $AB$ and $CD$ are two lines intersected by a transversal at points $x$ and $y$.

$$\begin{gathered} \angle AEP=50° \\ \angle CFQ=130° \end{gathered}$$

$\Rightarrow \angle EFD=\angle CFQ$ [$\because$,vertically opposite angles are equal]

$$\begin{gathered} \Rightarrow y=130°\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{\angle }\mathit{C}\mathit{F}\mathit{Q}\mathbf{=}\mathbf{130}\mathbf{°}\mathbf{]} \\ \Rightarrow y=130°\to \left(1\right) \end{gathered}$$

Now, $PQ$ is a straight line so, sum of all angles is $180°$.

$$\begin{gathered} \Rightarrow \angle AEP+\angle AEQ=180° \\ \Rightarrow 50°+x=180° \\ \Rightarrow x=180°-50° \\ \Rightarrow x=130°\to \left(2\right) \end{gathered}$$

Step-2 To show $AB\parallel CD$.

Now, From equation $\left(1\right)$ and $\left(2\right)$,

We say that $x=y$

As they are pair of alternate interior angles

So, $AB\parallel CD\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{\angle }\mathbf{x}\mathbf{=}\mathbf{\angle }\mathbf{y}\mathbf{]}$

Hence, the value of angles $x$ and $y$ is $130°$ and $AB\parallel CD$.