Example Question - parallel line segments
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Geometric Proof for Parallel Line Segments
To prove \( ABCD \parallel EFG \), we look for angle relationships that would imply parallelism through the Converse of the Corresponding Angles Postulate or the Converse of the Alternate Interior Angles Theorem. The correct statement that would allow Steve to prove \( ABCD \parallel EFG \) is: \( \angle CBF \cong \angle GFC \) Using the Converse of the Alternate Interior Angles Theorem, if \( \angle CBF \cong \angle GFC \), then line segment \( ABCD \) is parallel to line segment \( EFG \).
Finding the Value of x in Parallel Line Segments with Given Angle Expressions
\begin{align*} \text{Let } \angle ABC &= \angle BAC \text{ (alternate interior angles are equal because } AD \parallel BC) \\ 4x + 30^\circ &= 40^\circ \\ 4x &= 10^\circ \\ x &= \frac{10^\circ}{4} \\ x &= 2.5^\circ \\ \end{align*} \text{Round to the nearest tenth: } x = 2.5^\circ
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