f the bisectors of angles of a quadrilateral enclose a rectangle, then show that it is a parallelogram.

Sure, here is the step by step proof:

Step 1: Let's consider a quadrilateral ABCD. Let the bisectors of ∠A, ∠B, ∠C and ∠D meet at points E, F, G and H respectively.

Step 2: We are given that EFGH is a rectangle. This means ∠EFG = ∠FGH = ∠GHF = ∠HFE = 90°.

Step 3: Since EF and GH are opposite sides of rectangle EFGH, they are parallel and equal in length. Similarly, FG and HE are parallel and equal in length.

Step 4: Now, ∠AEB is half of ∠A and ∠BEF is half of ∠B. So, ∠AEB + ∠BEF = 1/2 ∠A + 1/2 ∠B = 1/2 (∠A + ∠B). But ∠A + ∠B = 180° (since opposite angles of a quadrilateral sum to 180°), so 1/2 (∠A + ∠B) = 90°. Therefore, ∠AEB + ∠BEF = 90°, which means line AB is parallel to EF.

Step 5: Similarly, we can prove that BC is parallel to FG, CD is parallel to GH, and DA is parallel to HE.

Step 6: Since both pairs of opposite sides of quadrilateral ABCD are parallel, ABCD is a parallelogram.

Therefore, if the bisectors of angles of a quadrilateral enclose a rectangle, then the quadrilateral is a parallelogram.