Prove that the bisectors of two angles of a parallelogram adjacent to one side are mutually perpendicular.

Let a parallelogram ABCD be given. By the property of a parallelogram, the angles adjacent to one side of the parallelogram add up to 180 °. If ∠ABC = α, ∠BCD = β, then α + β = 180 °. Let us draw the bisectors of the angles BК and СР, let them intersect at point M. Consider Δ ВСМ, in which ∠МВС = α / 2; ∠BCM = β / 2, then ∠ВМС = 180 ° – (∠МВС + ∠BCМ), since the sum of the angles in the triangle is 180 °. ∠ВМС = 180 ° – (α / 2 + β / 2) = 180 ° – (α + β) / 2 = 180 ° – (180 °) / 2 = 90 °. Hence, BK ⊥ CP. Since the angles are chosen arbitrarily, it can be concluded that the bisectors of the two angles of the parallelogram adjacent to one side are mutually perpendicular.
Q.E.D.