Internal bisectors of the angles of a parallelogram enclose a

Take parallelogram ABCD. Adjacent angles A and B are co-interior, so A + B = 180°. Their bisectors meet at an angle of (A/2 + B/2) = 90°. The same holds at every corner, so the figure enclosed by the four internal bisectors has four right angles. A quadrilateral with four right angles is a rectangle, so B is correct.

Why the others fail:

• Parallelogram: the enclosed figure has all right angles, which is a stricter condition than opposite angles being equal. Every rectangle is a parallelogram, but the converse does not hold here.
• Rhombus: a rhombus needs all four sides equal. A general parallelogram does not give equal sides for the enclosed figure, so it is not guaranteed.
• Square: a square needs equal sides as well as right angles. The right angles are guaranteed, but the equal sides are not.

The core concept is the angle sum property of co-interior angles along a transversal, combined with bisector geometry. It is basic Euclidean reasoning, useful for CSAT and for the geometry tested in reasoning sections.

Trap: aspirants see the symmetric look of the bisectors and pick rhombus or square. Check the angles first, then sides.