Question

given: \\( \overline { a b } \parallel \overline { c d } \\), \\( \overline { a b } \cong \overline { c d } \\), \\( \overline { a e } \cong \overline { c f } \\) prove: \\( \angle c b e \cong \angle a d f \\)

Explanation:

Step1: Prove \( \triangle ABE\cong\triangle CDF \)

Since \( \overline{AB}\parallel\overline{CD} \), then \( \angle BAE=\angle DCF \) (alternate - interior angles). Given \( \overline{AB}\cong\overline{CD} \) and \( \overline{AE}\cong\overline{CF} \), by the Side - Angle - Side (SAS) congruence criterion, \( \triangle ABE\cong\triangle CDF \). So, \( BE = DF \) and \( \angle AEB=\angle CFD \).

Step2: Prove \( \overline{BE}\parallel\overline{DF} \)

Since \( \angle AEB+\angle BEC = 180^{\circ} \) and \( \angle CFD+\angle DFA=180^{\circ} \), and \( \angle AEB=\angle CFD \), then \( \angle BEC=\angle DFA \). So, \( \overline{BE}\parallel\overline{DF} \) (alternate - interior angles are equal).

Step3: Prove \( \triangle BEC\cong\triangle DFA \)

We know \( \overline{BE}\parallel\overline{DF} \), so \( \angle EBC=\angle FDA \) (alternate - interior angles). Also, \( \overline{BE}\cong\overline{DF} \) (from Step 1) and \( \overline{BC}\cong\overline{DA} \) (because \( ABCD \) is a parallelogram, since \( \overline{AB}\parallel\overline{CD} \) and \( \overline{AB}\cong\overline{CD} \)). By the Side - Angle - Side (SAS) congruence criterion, \( \triangle BEC\cong\triangle DFA \).

Answer:

\( \angle CBE\cong\angle ADF \) (corresponding parts of congruent triangles \( \triangle BEC\) and \( \triangle DFA\) are congruent)