Consider parallel lines \(L1\) and \(L2\), intersected by transversal \(T\). Consecutive interior angles (on the same side of \(T\)) sum to \(180^\circ\). Let the angles be \(\alpha\) and \(\beta\). Their bisectors meet at an angle calculated thus: In the triangle formed by the bisectors and the transversal segment, the angles are \(\frac{\alpha}{2}\), \(\frac{\beta}{2}\), and the angle between bisectors (say \(\theta\)). Therefore, \(\frac{\alpha}{2} + \frac{\beta}{2} + \theta = 180^\circ\). Since \(\alpha + \beta = 180^\circ\), then \(\frac{180^\circ}{2} + \theta = 180^\circ\), meaning \(90^\circ + \theta = 180^\circ\), and \(\theta = 90^\circ\). The quadrilateral formed by the four interior angle bisectors has all angles equal to \(90^\circ\). A quadrilateral with four right angles is a Rectangle (3). It's a square only if the transversal is perpendicular to the parallel lines.