1. **Problem statement:** Given quadrilateral ABCD with AB = DC and \(\angle ABC = \angle BCD\) (both acute), prove that AD is parallel to BC. 2. **Key information:** - AB = DC (opposite sides equal) - \(\angle ABC = \angle BCD\) (equal acute angles) - Hint: Draw a line through A parallel to DC. 3. **Step 1: Construct line through A parallel to DC.** By the hint, draw line \(l\) through point A such that \(l \parallel DC\). 4. **Step 2: Analyze angles formed.** Since \(l \parallel DC\) and AB = DC, triangle ABB' (where B' is intersection of \(l\) and extension of BC) is isosceles with AB = DC = AB'. 5. **Step 3: Use equal angles.** Given \(\angle ABC = \angle BCD\), and since \(l \parallel DC\), alternate interior angles imply \(\angle B'AB = \angle BCD\). 6. **Step 4: Conclude AD is parallel to BC.** Since \(l\) passes through A and is parallel to DC, and \(\angle ABC = \angle BCD\), it follows that AD coincides with \(l\) and thus AD \(\parallel\) BC. **Final answer:** AD is parallel to BC.