A parallelogram has sides $a$ and $b$. What are the lengths of the diagonals of the quadrilateral formed by the angle bisectors of the parallelogram?
A parallelogram is centrally symmetric, therefore the angle bisectors of the opposite vertices are mirror images of each other, and thus parallel. Thus, the quadrilateral EFGH is also a parallelogram.

![](https://cdn.mathpix.com/cropped/2024_05_02_f2261a96e6ce6cbad252g-1.jpg?height=293&width=681&top_left_y=268&top_left_x=711)

Furthermore, \( \angle HAD + \angle HDA = \frac{1}{2}(\angle BAD + \angle ABC) = \frac{1}{2} \cdot 180^\circ = 90^\circ \), so \( \angle AHD = 180^\circ - (\angle HAD + \angle HDA) = 90^\circ \). Therefore, EFGH is a right-angled parallelogram, or a rectangle. The diagonals of a rectangle are equal in length, so it is sufficient to determine, for example, the length of the diagonal \( HF \).

The point \( H \) lies on both the angle bisectors of \( A \) and \( D \), so \( H \) is equidistant from the sides \( AB \) and \( AD \), as well as from the sides \( AD \) and \( DC \), and thus \( H \) lies on the midline of the parallelogram parallel to the sides \( AB \) and \( CD \). By symmetry, the same can be said about point \( F \), so \( HF \parallel CD \). Since \( HS \parallel FC \), \( HFCS \) is a parallelogram, \( HF = SC = DC - DS \). However, \( DS = DA \) because the triangle \( DAS \) is isosceles, since \( \angle DSA = \angle SAB \) (alternate interior angles), and \( \angle DAS = \angle SAB \) (since \( AS \) is the angle bisector). Therefore, \( HF = DC - DA \), and thus the diagonals of the quadrilateral EFGH are equal to the difference of the lengths of two adjacent sides of the parallelogram \( ABCD \), \( |a - b| \).