Yes. If the diagonals bisect the angles, the quadrilateral is always a parallelogram, specifically, a rhombus.
Consider quadrilateral ABCD. If diagonal AC bisects angles A and C, then ΔACB is congruent to ΔACD (ASA). Hence AB=AD and BC=CD (CPCTC).
Likewise, if diagonal BD bisects angles B and D, triangles BDA and BDC are congruent, thus AB=BC and AD=CD. (CPCTC again). Now, we have AB=BC=CD=AD, so the figure is a rhombus, hence a parallelogram.
