Prove that the angle bisectors of a parallelogram form a rectangle.

Sure, here is the step-by-step proof that the angle bisectors of a parallelogram form a rectangle:

Step 1: Start with a parallelogram ABCD. Draw the angle bisectors of ∠A, ∠B, ∠C, and ∠D. These bisectors will intersect at points E, F, G, and H, forming a quadrilateral EFGH.

Step 2: We know that in a parallelogram, opposite angles are equal. Therefore, ∠A = ∠C and ∠B = ∠D.

Step 3: Since E, F, G, and H are the intersections of the angle bisectors, ∠AEF = ∠CFH and ∠BEH = ∠DFG. This is because the angles are half of ∠A, ∠B, ∠C, and ∠D respectively.

Step 4: Now, we have two pairs of opposite angles in quadrilateral EFGH that are equal, which means EFGH is a cyclic quadrilateral.

Step 5: In a cyclic quadrilateral, the sum of a pair of opposite angles is 180 degrees. Therefore, ∠EHF + ∠EFG = 180 and ∠EFG + ∠EGH = 180.

Step 6: From step 5, we can conclude that ∠EHF = ∠EGH. Similarly, we can prove that ∠HEF = ∠HGF.

Step 7: Since we have two pairs of adjacent angles that are equal, EFGH is a rectangle.

Therefore, the angle bisectors of a parallelogram form a rectangle.