Question

given: \\( \overline { b c } \parallel \overline { a d } \\) and \\( \overline { a b } \parallel \overline { c d } \\).
prove: \\( \angle b c d \cong \angle b a d \\).

Explanation:

Step1: Use alternate - interior angles

Since $\overline{BC}\parallel\overline{AD}$ and $\overline{AC}$ is a transversal, $\angle BCA\cong\angle DAC$ (alternate - interior angles).
Since $\overline{AB}\parallel\overline{CD}$ and $\overline{AC}$ is a transversal, $\angle BAC\cong\angle DCA$ (alternate - interior angles).

Step2: Use angle - addition property

$\angle BCD=\angle BCA+\angle DCA$ and $\angle BAD = \angle BAC+\angle DAC$.
By substitution (since $\angle BCA\cong\angle DAC$ and $\angle BAC\cong\angle DCA$), we have $\angle BCD\cong\angle BAD$.

Answer:

$\angle BCD\cong\angle BAD$