91Ó°ÊÓ

Arrowheads denotes line segment $\stackrel{¯}{AB}$is parallel to $\stackrel{¯}{CD}$

Double arrowheads denotes line segment $\stackrel{¯}{BC}$is parallel torole="math" localid="1637750713301" $\stackrel{¯}{DE}$

Line segments $\stackrel{¯}{BC}\text{and}\stackrel{¯}{XY}$are transversal to the parallel lines $\stackrel{¯}{AB}$and$\stackrel{¯}{CD}$

Line segments $\stackrel{¯}{CD}\text{and}\stackrel{¯}{XY}$are transversal to the parallel lines $\stackrel{¯}{BC}$and$\stackrel{¯}{DE}$

$∠ABC$and $∠BCD$forms a pair of alternate interior angles

$∠BCD$and $∠CDE$forms another pair of alternate interior angles

Theorem 3-2: If two parallel lines are cut by a transversal then the alternate interior angles are congruent.

$\begin{array}{c}∠BCDâ‰\dots ∠ABC\end{array}$

$m∠BCD=m∠ABC$

$m∠BCD=55$

$∠CDEâ‰\dots ∠BCD$

$m∠CDE=m∠BCD$

$x=55$

$∠ACB$and $∠CED$forms a pair of corresponding angles

$∠BAC$and $∠DCE$forms a pair of corresponding angles

Postulate 10: If two parallel lines are cut by a transversal then the corresponding angles are congruent

$∠ACBâ‰\dots ∠CED$

$m∠ACB=m∠CED$

$m∠ACB=50$

$∠DCEâ‰\dots ∠BAC$

$m∠DCE=m∠BAC$

$m∠DCE=y$

Since, $∠ACB$, $∠BCD$and $∠DCE$forms the straight line $\stackrel{¯}{AE}$ so these angles are supplementary angles. That is,

$m∠ACB+m∠BCD+m∠DCE=180$

$50+55+y=180$

$105+y=180$

$y=180$

$y=75$

Thus $y=75$