The quadrilateral so formed is a rhombus (a parallelogram in which all the sides are of equal length). The interior angles of the rhombus are 90+x and 90-x where x is the angle of the parallelogram and where 90+x are the measures of opposite angles of the rhombus. The adjacent angles add up to 180, and 90+x+90-x=180.
Call the parallelogram ABCD and let angle BAD be x. Because opposite side and opposite angles of a parallelogram are equal, we only need to consider two sides. Let AB=a and AD=b be two adjacent sides. Let the bisectors of these sides be points P and Q respectively. AP=PB=a/2 and AQ=QD=b/2. Outside ABCD, draw XP, length a/2, perpendicular to AB, and YQ, length b/2, perpendicular to AD. X and Y are the centres of the squares on AB and AD.
One side of the new quadrilateral is defined by the line XY. So let's find its length.
AX=a/sqrt(2) and AY=b/sqrt(2), because they are hypotenuses of triangles AXP and AYQ. XY=a^2/2+b^2/2-abcosXAY (cosine rule). Angle XAY=XAP+BAD+QAY=45+x+45=90+x. cosXAY=cos(90+x)=-sinx. So XY=a^2/2+b^2/2+absinx.
Let's find the length of an adjacent side of the new quadrilateral. This time we have angle CDA=180-x. CD is bisected at R and WR is the perpendicular on to CD, and WR=RD=RC=a/2 because DC is parallel and equal in length to AB. Angle YDW=360-(RDW+180-x+QDY)=360-(45+180-x+45)=90+x, and cos(90+x)=-sinx. WY=DW^2+DY^2-2DWDYcosYDW=a^2/2+b^2/2+absinx. Therefore WY=XY. Since there is symmetry in the geometry the other two sides will be the same. And the opposite interior angles are equal. Therefore we have rhombus.
When you draw the diagram you may find that XY is exterior to the parallelogram, and WY is interior. Nevertheless, since sin(180-x)=sinx, the lengths of the sides will still be equal. There are various other geometries that may apply, but you will find that you always get a combination of two angles of 45 degrees and the angle of the parallelogram (skew), but sinx will always be the result (for example, cos(90-x)=sinx, cos(45-x+45)=sin(x), cos(180-(90+x))=sinx, etc.).