Prove that the bisector of the inner angle of a parallelogram cuts off an isosceles triangle from the parallelogram.

Prove that the bisector of the inner angle of a parallelogram cuts off an isosceles triangle from the parallelogram. Can this triangle also be equilateral?

Let a parallelogram ABCD be given: AK is the bisector ∠A, thus ∠BAK = ∠DAK = ∠A / 2.
Since lines AD and BC are parallel, line AK is a secant, intersecting two parallel lines, then ∠DAK = ∠BKA as criss-crossing angles.
In △ ABK ∠BAK = ∠BKA, therefore, △ ABK is an isosceles triangle with sides BA and BK and base AK, as required.
2. A triangle cut off from the parallelogram by the bisector of the inner angle can be equilateral (regular) if the bisector is omitted from an angle of 120 °.
Let ∠A = 120 °, then:
∠BAK = ∠DAK = ∠A / 2 = 120 ° / 2 = 60 °.
∠DAK = ∠BKA = 60 ° as criss-cross.
B △ ABK ∠BAK = ∠BKA = 60 °.
By the theorem on the sum of the angles of a triangle:
∠BAK + ∠ABK + ∠BKA = 180 °;
60 ° + ∠ABK + 60 ° = 180 °;
∠ABK = 180 ° – 120 °;
∠ABK = 60 °.
Thus, all angles in все ABK are equal to 60 °, which means that it is equilateral.